Binomial theorem

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Binomial theorem

  1. 1. Binomial Theorem The theorem is called binomial because it is concerned with a sum of two numbers (bi means two) raised to a power. Where the sum involves more than two numbers, the theorem is called the Multi-nomial Theorem. The Binomial Theorem was first discovered by Sir Isaac Newton. Exponents of (a+b) Now on to the binomial. We will use the simple binomial a+b, but it could be any binomial. Let us start with an exponent of 0 and build upwards. Exponent of 0 When an exponent is 0, you get 1: (a+b)0 = 1 Exponent of 1 When the exponent is 1, you get the original value, unchanged: (a+b)1 = a+b Exponent of 2 An exponent of 2 means to multiply by itself (see how to multiply polynomials): (a+b)2 = (a+b)(a+b) = a2 + 2ab + b2 Exponent of 3 For an exponent of 3 just multiply again: (a+b)3 = (a+b)(a2 + 2ab + b2) = a3 + 3a2b + 3ab2 + b3
  2. 2. We have enough now to start talking about the pattern. The Pattern In the last result we got: a3 + 3a2b + 3ab2 + b3Now, notice the exponents of a. They start at 3 and go down: 3, 2, 1, 0: Likewise the exponents of b go upwards: 0, 1, 2, 3: If we number the terms 0 to n, we get this: k=0 k=1 k=2 k=3 a3 a2 a 1 1 b b2 b3 Which can be brought together into this: an-kbk How about an example to see how it works: Example: When the exponent, n, is 3. The terms are: k=0: k=1: k=2: k=3: an-kbk an-kbk an-kbk an-kbk = a3-0b0 = a3-1b1 = a3-2b2 = a3-3b3
  3. 3. = a3 = a2b = ab2 = b3 It works like magic! Coefficients So far we have: a3 + a2b + ab2 + b3 But we really need: a3 + 3a2b + 3ab2 + b3 We are missing the numbers (which are called coefficients).Lets look at all the results we got before, from (a+b)0 up to (a+b)3:And now look at just the coefficients (with a "1" where a coefficient wasnt shown):
  4. 4. They actually make Pascals Triangle!Each number is just the two numbersabove it added together (except for theedges, which are all "1") (Here I have highlighted that 1+3 = 4)Armed with this information let us try something new ... an exponent of 4: a exponents go 4,3,2,1,0: a4 + a3 + a2 + a + 1 b exponents go 0,1,2,3,4: a4 + a3b + a2b2 + ab3 + b4 coefficients go 1,4,6,4,1: a4 + 4a3b + 6a2b2 + 4ab3 + b4 And that is the correct answer. We have success!We can now use that pattern for exponents of 5, 6, 7, ... 50, ... 112, ... you name it! As a Formula Our last step is to write it all as a formula.But hang on, how do we write a formula for "find the coefficient from Pascals Triangle" ... ? Well, there is such a formula: It is commonly called "n choose k" because it is how many ways to choose k elements from a set of n. You can read more at Combinations and
  5. 5. Permutations The "!" means "factorial", for example 4! = 1×2×3×4 = 24 Example: Row 4, term 2 in Pascals Triangle is "6". Lets see if the formula works: Yes. Correct. Putting It All Together The last step is to put all the terms together into one formula.But we are adding lots of terms together ... can that be done using one formula? Yes! The handy Sigma Notation allows us to sum up as many terms as we want: Sigma Notation Now it can all go into one formula: The Binomial Theorem Use It
  6. 6. OK ... it wont make much sense without an example. So lets try using it for n = 3 : BUT ... it is usually much easier just to remember the patterns: The first terms exponents start at n and go down The second terms exponents start at 0 and go up Coefficients are from Pascals Triangle, or by calculation using n!/(k!(n-k)!) Like this: Example: What is (x+5)4 Start with exponents: x450 x351 x252 x153 x054 Include Coefficients: 1x450 4x351 6x252 4x153 1x054Then write down the answer (including all calculations, such as 4×5, 6×52, etc): (x+5)4 = x4 + 20x3 + 150x2 + 500x + 625 You may also want to calculate just one term: Example: What is the coefficient for x3 in (2x+4)8
  7. 7. The exponents for x3 are: (2x)345 The coefficient is "8 choose 3". We can use Pascals Triangle, or calculate directly: n! 8! 8! 8×7×6 = = = = 56 k!(n-k)! 3!(8-3)! 3!5! 3×2×1 And we get: 56(2x)345 Which simplifies to: 458752 x3 A large coefficient, isnt it? And one last, most amazing example: Example: A formula for e (Eulers Number) You can use the Binomial Theorem to calculate e (Eulers number).e = 2.718281828459045... (the digits go on forever without repeating) It can be calculated using: (1 + 1/n)n (It gets more accurate the higher the value of n)That formula is a binomial, right? So lets use the Binomial Theorem:
  8. 8. First, we can drop 1n-k as it is always equal to 1:And, quite magically, most of what is left goes to 1 as n goes to infinity: Which just leaves: With just those first few terms we get e ≈ 2.7083...
  9. 9. The Binomial Theorem is a quick way (okay, its a less slowway) of expanding (or multiplying out) a binomial expressionthat has been raised to some (generally inconveniently large)power. For instance, the expression (3x – 2)10 would be verypainful to multiply out by hand. Thankfully, somebody figuredout a formula for this expansion, and we can plug the binomial3x – 2 and the power 10 into that formula to get that expanded(multiplied-out) form.The formal expression of the Binomial Theorem is as follows: Copyright © Elizabeth Stapel 1999-2009 All Rights ReservedWhereRecall that the factorial notation "n!" means " the product of allthe whole numbers between 1 and n", so, for instance, 6! =1×2×3×4×5×6. Then the notation "10C7" (often pronounced as"ten, choose seven") means:There is another way to find the value of "nCr", andits called "Pascals Triangle". To make thetriangle, you start with a pyramid of three 1s, likethis:Then you get the next row of numbers by addingthe pairs of numbers from above. (Where there isonly one number above, you just carry down the
  10. 10. 1.)Keep going, alwaysadding pairs of numbersfrom the previous row..To find, say, 6C4, you godown to the row wherethere is a "6" after theinitial "1", and then goover to the 5th (notthe 4th) entry, to find that6C4 = 15.As you might imagine, drawing Pascals Triangle every timeyou have to expand a binomial would be a rather long process,especially if the binomial has a large exponent on it. Peoplehave done a lot of studies on Pascals Triangle, but in practicalterms, its probably best to just use your calculator to find nCr,rather than using the Triangle. The Triangle is cute, I suppose,but its not terribly helpful in this context, being more time-consuming than anything else. For instance, on a test, do youwant to evaluate "10C7" by calculating eleven rows of theTriangle, or by pushing four buttons on your calculator?I could never remember the formula for the Binomial Theorem,so instead, I just learned how it worked. I noticed that thepowers on each term in the expansion always added up towhatever n was, and that the terms counted up from zero to n.Returning to our intial example of (3x – 2)10, the powers onevery term of the expansion will add up to 10, and the powerson the terms will increment by counting up from zero to 10:
  11. 11. (3x – 2)10 = 10C0 (3x)10–0(–2)0 + 10C1 (3x)10–1(–2)1 + 10C2 (3x)10– 2 (–2)2 + 10C3 (3x)10–3(–2)3 + 10C4 (3x)10–4(–2)4 + 10C5 (3x)10–5(–2)5 + 10C6 (3x)10–6(–2)6 + 10C7 (3x)10–7(–2)7 + 10C8 (3x)10–8(–2)8 + 10C9 (3x)10–9(–2)9 + 10C10 (3x)10–10(–2)10Note how the highlighted counter number counts up from zeroto 10, with the factors on the ends of each term having thecounter number, and the factor in the middle having the counternumber subtracted from 10. This pattern is all you really need toknow about the Binomial Theorem; this pattern is how it works. Expand (x2 + 3)6 Students trying to do this expansion in their heads tend to mess up the powers. But this isnt the time to worry about that square on the x. I need to start my answer by plugging the terms and power into the Theorem. The first term in the binomial is "x2", the second term in "3", and the power n is 6, so, counting from 0 to 6, the Binomial Theorem gives me: (x2 + 3)6 = 6C0 (x2)6(3)0 + 6C1 (x2)5(3)1 + 6C2 (x2)4(3)2 + 6C3 (x2)3(3)3 + 6C4 (x2)2(3)4 + 6C5 (x2)1(3)5 + 6C6 (x2)0(3)6 Then simplifying gives me (1)(x12)(1) + (6)(x10)(3) + (15)(x8)(9) + (20)(x6)(27) + (15)(x4)(81) + (6)(x2)(243) + (1)(1)(729) = x12 + 18x10 + 135x8 + 540x6 + 1215x4 + 1458x2 + 729
  12. 12. Expand (2x – 5y)7 Ill plug "2x", "–5y", and "7" into the Binomial Theorem, counting up from zero to seven to get each term. (I mustnt forget the "minus" sign that goes with the second term in the binomial.) (2x – 5y)7 = 7C0 (2x)7(–5y)0 + 7C1 (2x)6(–5y)1 + 7C2 (2x)5(– 5y)2 + 7C3 (2x)4(–5y)3 + 7C4 (2x)3(–5y)4 + 7C5 (2x)2(–5y)5 + 7C6 (2x)1(–5y)6 + 7C7 (2x)0(–5y)7 Then simplifying gives me: Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved (1)(128x7)(1) + (7)(64x6)(–5y) + (21)(32x5)(25y2) + (35)(16x4)(–125y3) + (35)(8x3)(625y4) + (21)(4x2)(–3125y5) + (7)(2x)(15625y6) + (1)(1)(–78125y7) = 128x7 – 2240x6y + 16800x5y2 – 70000x4y3 + 175000x3y4 – 262500x2y5 + 218750xy6 – 78125y7You may be asked to find a certain term in an expansion, theidea being that the exercise will be way easy if youvememorized the Theorem, but will be difficult or impossible if youhavent. So memorize the Theorem and get the easy points. What is the fourth term in the expansion of (3x – 2)10? Ive already expanded this binomial, so lets take a look: (3x – 2)10 = 10C0 (3x)10–0(–2)0 + 10C1 (3x)10–1(–2)1 + 10C2 (3x)10–2(–2)2
  13. 13. + 10C3 (3x)10–3(–2)3 + 10C4 (3x)10–4(–2)4 + 10C5 (3x)10– 5 (–2)5 + 10C6 (3x)10–6(–2)6 + 10C7 (3x)10–7(–2)7 + 10C8 (3x)10– 8 (–2)8 + 10C9 (3x)10–9(–2)9 + 10C10 (3x)10–10(–2)10So the fourth term is not the one where Ive counted up to4, but the one where Ive counted up just to 3. (This isbecause, just as with Java script, the counting starts with0, not 1.)Note that, in any expansion, there is one more term thanthe number in the power. For instance: (x + y)2 = x2 + 2xy + y2 (second power: three terms) (x + y)3 = x3 + 3x2y + 3xy2 + y3 (third power: four terms) (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 (fourth power: five terms)The expansion in this exercise, (3x – 2)10, has power of n =10, so the expansion will have eleven terms, and the termswill count up, not from 1 to 10 or from 1 to 11, but from 0 to10. This is why the fourth term will not the one where Imusing "4" as my counter, but will be the one where Imusing "3". 10 C3 (3x)10–3(–2)3 = (120)(2187)(x7)(–8) = –2099520x7Find the tenth term in the expansion of (x + 3)12.To find the tenth term, I plug x, 3, and 12 into the BinomialTheorem, using the number 10 – 1 = 9 as my counter: 12 C9 (x)12–9(3)9 = (220)x3(19683) = 4330260x3Find the middle term in the expansion of (4x – y)8.
  14. 14. Since this binomial is to the power 8, there will be nine terms in the expansion, which makes the fifth term the middle one. So Ill plug 4x, –y, and 8 into the Binomial Theorem, using the number 5 – 1 = 4 as my counter. 8 C4 (4x)8–4(–y)4 = (70)(256x4)(y4) = 17920x4y4You might be asked to work backwards. Express 1296x12 – 4320x9y2 + 5400x6y4 – 3000x3y6 + 625y8 in the form (a + b)n. I know that the first term is of the form an, because, for whatever n is, the first term is nC0 (which always equals 1) times an times b0 (which also equals 1). So 1296x12 = an. By the same reasoning, the last term is bn, so 625y8 = bn. And since there are alternating "plus" and "minus" signs, I know from experience that the sign in the middle has to be a "minus". (If all the signs had been "plusses", then the middle sign would have been a "plus" also. But in this case, Im really looking for "(a – b)n".) I know that, for any power n, the expansion has n + 1 terms. Since this has 5 terms, this tells me that n = 4. So to find a and b, I only have to take the 4th root of the first and last terms of the expanded polynomial: Then a = 6x3, b = 5y2, there is a "minus" sign in the middle, and: 1296x12 – 4320x9y2 + 5400x6y4 – 3000x3y6 + 625y8 = (6x3 – 5y2)4Dont let the Binomial Theorem scare you. Its just anotherformula to memorize. A really complicated and annoying
  15. 15. formula, Ill grant you, but just a formula, nonetheless. Dontoverthink the Theorem; there is nothing deep or meaningfulhere. Just memorize it, and move onBinomial Theorem for positive integral index: (x+y)n = xn + nC1xn-1y+nC2xn-2y2+-----+nCrxn-ryr+ -------+---------+nCn- n-11xy + ncnyn.It can be represented as:(x+y)n = n Crxn-ryrParticular – Cases :(i) Replacing ‘y’ by ‘-y’, we have : (x-y)n = nCoxyo-nC1xn-1y+nC2xn-2y2-------+(-1)r nCrxn-ryr+------+(- 1)n nCnxoyn.It can be represented as : (x+y)n = (-1)r nCrxn-ryr(ii) Replacing ‘x’ by ‘1’ and ‘y’ by ‘x’, we have :(1+x)n = nCoxo+nC1x+nC2x2+---------+nCrxr+------+nCn-1xn-1+nCnxn. n or = Crxr(ii) Replacing ‘x’ by ‘-x’, we have :(1+x)n = nCoxo-nC1x1+nC2x2 - ---------+(-1)r nCrxr+------+nCn-1(-1) n-1 +(-1)n nCnxn.or = (-1)rnCrxr
  16. 16. Properties of Binomial – Expansion (x+y)n : (i) There are (n+1) terms in the expansion. (ii) In each term, sum of the indices of ‘x’ and ‘y’ is equal to ‘n’. (iii) In any term, the lower suffix of ‘c’ is equal to the index of ‘y’, and the index of x = n-(lower suffix of c). (iv) Because nCr = nCn-r, so we have : n Co = nCn n C1=nCn-1 n C2=nCn-2 etc. It follows that the coefficients of terms equidistant from the beginning and the ends are equalQ2: In the expansion of (x+a)n, if the sum of odd-terms be ‘P’ and sum of even be ‘Q’ Prove that: (i) P2-Q2 = (x2-a2)n (ii) 4PQ = (x+a)2n – (x-a)2nSol.: (x+a)n = xn+nC1xn-1a + nC2xn-2a2+nC3xn-3a3 + ------- + nCnan = (xn+nC2xn-2a2 + ---------) + (nC1xn-1a+nC3xn-3a3+ ------) (x+a)n = P+Q ------------------------> (1)and (x-a)n = xn - nC1xn-1a+nC2xn-2a2-nC3xn-3a3 + ----+ (-1)n nCnan = (xn+nC2xn-2a2 + ----) – (nC1xn-1a+nC3xn-3a3+------) (x-a)n = P – Q ----------------> (2)
  17. 17. Now we have :(1) P2 – Q2 = (P+Q) (P-Q) = (x+a)n (x-a)n = P2 – Q2 = (x2-a2)n(2) 4 PQ = (P+Q)2 – (P-Q)2 = 4 PQ = (x+a)2n – (x-a)2n**Q4. Prove that (101)50 > (100)50 + (99)50Sol.: (101)50 = (100+1)50 = (100)50 + 50c1(100)49 + 50c2(100)48 + -------+1 ------ (i) (99)50 = (100-1)50 = (100)50 – 50c1(100)49 + 50c2(100)48 - --------+1 ------(ii)eq(i) – eq(ii) :(101)50 – (99)50 = 2[50C1(100)49 + 50C3(100)47 + --------] = 2 x ( 50!/ 1! 49!) (100)49 + 2. 50 C3 47(100) + ------- = 100 (100)49 + (A positive number) = (100)50 + (A positive number)(101)50 – (99)50 > (100)50or (101)50 > (101)50 + (99)50General Terms : (r +1) th term from beginning in
  18. 18. (x+y)n is called general – term, andit is denoted by Tr+1 = nCrxn-ryrExplanation: We know(x-y)n = nCoxnyo+nC1xn-1y1+nC2xn-2y2+----+nCnxoyoHere:First term T1 = nCoxnyo T2 = nC1xn-1y1 T3 = nC2xn-2y2 ------------------------ ------------------------ ------------------------ Tr = nCr-1 xn-(r-1) yr-1Putting r = r+1 in this expression, we get:General Term: Tr+1 = nCr xn-r yrNote : ‘Tr’ can be used as general terms also. Problem based on General TermsType : 1 Find the 7th term in the expansion of[4x – (1 / 2 x)]13 13Sol : T7 = T6+1 = C6(4x)13-6 - (1/2 x) 6 13 = C6.47x7. 1 /(26.x3)
  19. 19. 13 = C6. 28.x4 = 13!/ (6!x7!) . 28. x4 = T7 = 439296x4Type II Find the coefficient of x-7 in the expansion of ( ax- )11Sol GeneralTerm,Tr+1 = 11Cr(ax)11-r - (1/bx2) rTr+1 =(-1)r11Cr.(a11-r/br) x11-3r ------------- (i)Putting 11 – 3r = -7Or 3r = 18 r=6From (i) to T7 = (-1)6 11C6. ( a5 / b6) x-7 -------------- (ii)Hence, the coefficient of x-7 in ax- (1 / b x2) 11 is 11C6a5b-6Type III : Find the term independent of ‘x’ in [(3 x2 / 2) – (1/ 3x) ] 9Sol.: General Term, Tr+1 = 9Cr (3 x2 / 2) 9-r – (1/3x) r = (-1)r 9Cr ( 3/2) 9-r x18-2r (1 / 3r. xr ) Tr+1 = (-1)r9 Cr (39-2r / 29-r). x18-3r ------- (i) Putting 18- 3 r = o r = 6So, from (i), 7th term is independent of ‘x’, and its value is:T7 = (-1)6 . 9C6. (3-3 / 23) xo
  20. 20. = 9 ! /(6! 3!) . 1/ (33 2 3)= T7 = (7/18)Pth term from end:‘P’th term from end in the expansion of (x+y)n is (n-P+2)th term frombeginning.Ex.: Find the 4th term from the end in the expansion of [ - ]7Sol.: 4th term from end = (7-4+2)th or 5th term from beginning.T5 = T4+1 = 7C4 (x3/2)7-4 . (-2/x4) 4 = 7C4 (x3 /2) 3 ( -2/ x2) 4 = 7! / (4! 3!) . (x9/8) . (16/ x8) = (7.6.5 / 3.2.1) .2xT5 = 70xHence ‘4’ term, from the end = 70x.Middle Terms: It depends upon the value of ‘n’.Case -1 : When ‘n’ is even, then total number of terms in (x+y) n isodd. So there is only one middle term i.e. [(n/2) + 1] th them is themiddle term.So we find (Tn+1/2). th term in this case, if ‘n’ is even.
  21. 21. Case II : When ‘n’ is odd, then total number of terms in (x+y)n iseven. So there are two middle terms i.e. (n+1) /2 th and (n+3) /2 thare true middle terms.so we find T(n+1)/2 th and T(n+3)/2 th in this case if ‘n’ is odd.Ex.: Find the middle – term in the expansion of [ 3x – ]9Sol.: Here total no. of terms are 10 (even). So there are true middle-termsi.e (9+1) / 2 th and (9+3) / 2 th. So we have to find – out ‘T5’ and‘T6’. T5 = T4+1 = 9C4(3x)9-4 (-x3 / 6) 4 = 9! / (4! 5!) .35 x5 ( x12 / 64)= (9.8.7.6 / 4.3.2.1) 35 / (24 x 34) x17 T5 = (189 / 8) x17 T6 = T5+1 = 9C5(3x)9-5 (-x3 / 6) 5= 9! / (5! 4!) .34 x4 (x15 / 65)= -(9.8.7.6 / 4.3.2.1) 34(25 x 35) x19 T6 = - ( 21 / 16) x19Greatest – term in (1+x)n : If ‘Tr’ and ‘Tr+1’ be the ‘r’ th and (r+1)thterms in theExpansion of (1+x)n, then : Tr+1 = nCr(1)n-r xr = nCr xr
  22. 22. And Tr = nCr-1. xr-1So: Tr+1 / Tr = (nCr xr / nCr-1 xr-1) = (n-r+1)/r |x|If ‘Tr+1 be the greatest term, then Tr+1 > TrOr Tr+1 / Tr > 1**Question: Find numerically the greatest term in the expansion of (2+3x)9 when x = (3 / 2)Sol.: 1 Method : (2+3x)9 = 29 [1+ 3x / 2] 9In the expansion of [(1 + 3x) / 2] 9, we have : Tr+1 / Tr = (9-r+1)/ r |3x / 2| 3 = ((10 – r)/r) | (3/2) x(3/2) |= (10 – r) / r x 9/4Tr+1 / Tr = (90- 9r) / 4rPutting Tr+1 / Tr ≥ 1 (90-9r) / 4r ≥ 1 or 90 ≥ 13 ror r 90 / 13or r ≤ 6 + 12 / 13
  23. 23. T6+1 or ‘T7’ is the greatest term.‘T7’ in [1 + (3x / 2)] 9 9T7 = T6+1 = C6 (3x / 6) 6= 9! / (3! 6!) .[ (3 / 2) (3 / 2)] 6 = (9 .8.7 / 3.2.1) (96 / 46) = (3 7 96) / 45 = (3 7 312) / 210 = 7. (313 / 210)So greatest term in (2+ 3x)9 is : = 29. 7. (313 / 210) = (7 313) / 2II- Method : (2+3x)9 = 29 [(1 + 3x) / 2] 9= 29 [1 + 9 / 4] 9since x = 3 / 2Here m = | x (n + 1) / (x + 1)| = | 9/4 (9+1) / 9/4 + 1| = 90 / 13So greatest term in the expansion is T[m]+1 = T3+1 = T7Now the method is same as in method (1)Greatest Coefficient : In any binomial expansion middle-term has thegreatest.Coefficient. So (i) If ‘n’ is even, then greatest – coefficient = nCn/2
  24. 24. (ii) If ‘n’ is odd, then greatest – coefficients are nC(n+1)/ 2 and n C (n-1)/2Properties of Binomial coefficients :(1) The sum of binomial coefficient in (1 + x)n is 2n.Proof (1 + x)n = Co+C1x+C2x2 + ----- + Cnxn----------- (i)Putting x = 1 :2n = Co + C1 + C2 + ----------- + Cn ----------- (ii)Ex.: Prove that the sum of the coefficients in the expression (1+x – 3x2)2163 is ‘-1’.Sol.: Putting x = 1 in (1 + x – 3x2)2163Some of the coefficients = (1 + 1 – 3)2163 = (-1)2163 = -1(2) The sum of the coefficients of the odd-terms in (1+x)n is equal tothe sum of coefficients of the even terms and each is equal to 2n-1.Proof: Putting x = -1, in eg(1) :O = Co – C1 + C2 – C3 + ------ + (-1)nCnand from (ii): 2n = Co + C1 + C2 + --------- + CnAdding these egn:2n = 2 ( Co + C2 + C4 + ---------------)or Co + C2 + C4 + ------- = 2n-1 ------------> (ii)Subtracting these egn:
  25. 25. 2n = 2 (C1 + C3 + C5 + --------------)or C1 + C3 + C5 + ------- = 2n-1 ------------> (iv)From (iii) and (iv) :C0 + C2 + C4 + ------- = C1 + C3 + C5 + ------- = 2n-1Ex.: Evaluate the sum of the 8C1 + 8C3 + 8C5 + 8C7Sol.: since nC1 + nC3 + nC5 + nC7 + -------- = 2n-1Here n = 8 8 C1 + 8C3 + 8C5 + 8C7 = (28-1)= 27= 128( 8C9, 8C11 etc. are not possible)Some important results: (i) In the expansion of (1+x)n, coefficient of xr = nCr (ii) In the expansion of (1-x)n, coefficient of xr = (-1)r. nCr (iii) If ‘n’ is a negative integer or fraction, then (1+x)n = 1 + (n / 1!) x + [ n (n-1)/ 2!] x2 + [n(n-1)(n-2) / 3!] x3 + ------------- + [n(n-1)(n-2) -----------(n-r+1) / r!]xr + -------------- Here | x | <1, i.e. – 1<x<1 is necessary for its validity. (iv) In (1+x)n, general – term Tr+1 = [n(n-1)(n-2) ------------ -(n-r+1) / r!]. x2 n (v) Cr + nCr-1 = n+1Cr
  26. 26. n (vi) Cx = nCy x = y or x + y = n n n-1 (vii) Cr = n/ r. Cr-1**Multinomial theorem : (For a ‘+’ve integral index):If n N, and x1, x2, x3, --------xm C, then(x1 + x2 + x3 + ---------+xm)n = ? n! / (n1! n2! ---nm!) x1n1, x2n2….xmnmWhere n1, n2, n3 --------, nm are non-negative integers, satisfying theconditionn1 + n2 + -----------+nm = nNote: The coefficient of x1n1. x2n2. ---------xmnm in the expansion of(x1 + x2 + x3 + ------------------- + xm)n is := n! / (n1! x n2! ---nm!)So, general-term in (a+b+c+d)n = n! / (p! x q! x r! x s!).ap.bq.cr.ds.Where p+q+r+s = n, and p, q, r, s W.(2) Number of terms in (x1 + x2 + x3 + --------- + xm)n : n+m-1 Cm-1.Ex.: Find the number of terms in the expansion of (2x – 3y + 4z)100 100+3-1Sol.: Number of terms = C3-1 = 102C2 = 102 ! / (2! X 100!) = (102 x 101) / (2 x 1) = 5151**General term of a multinomial – theorem :
  27. 27. Tr+1 = n! / (n1! x n2! ---nm!) x1n1. x2n2 -----------xmnm EXAMPLES**Q1. Find the coefficient of x3 y4 z2 in the expansion of (2x – 3y + 4z)9.Sol. General Term in (2x – 3y + 4z)9= 9! / (n1! n 2! n3!). (2x)n1. (-3y)n2. (4z)n3= 9! / (n1! n 2! n3!). 2n1 (-3)n2. (4)n3. xn1. yn2. zn3Putting n1 = 3, n2 = 4, n3 = 2 : = 9! / (3! 4! 2!). 23 (-3)4. (4)2. x3 y4 z2 = [ 9. 8. 7. 6. 5. 4!/ (3.2.1. 4!.2)] x 8 81 16 x3 y4 z2 Coefficient of x3 y4 z2 = 9 . 8 . 7 . 5 . 8 . 81 . 8= 13063600Greatest coefficient in the expansion of (x1 + x2 + -------- + xm)n is= n! / (q!) m-r ( q+1!) rWhere ‘q’ is the quotient and ‘r’ is the remainder, when ‘n’ is dividedby ‘m’.**Example: Find the greatest coefficient in the expansion of (a + b + c + d) 15.Sol.: Here n = 15, m = 415/4 is quotient 3 and remainder 3.since q = 3 and r = 3
  28. 28. Hence greatest – coefficient = 15! / [(3!) 4-3 x (3+1!)3]= 15! / [(3!) x (4!)3 ]= 15! / (3! x 4! x 4! x 4!)Ex.: Find the coefficient of x7 in the expansion of (1+3x-2x3)10.Sol.: General term in (1+3x-2x3)10= 10! / (n1! x n2! x n3!). (1)n1 (3x)n2 (-2x3)n3= 10! / (n1! x n2! x n3!). 3n2 (-2)n3 xn2+3n3Where n1 + n2 + n3 = 10 --------------> (i)For coefficient of x7 : n2 + 3n3 = 7 -------------> (ii)From (ii), possible non-negative integral values of ‘n2’ and ‘n3’ are : n2 = 7, n3 = o since from (i) : n1 = 3 n2 = 1, n3 = 2 since from(i) : n1 = 7or n2 = 4, n3 = 1 since from (i): n1 = 5So required coefficient of x7 : 10! / (3! 7! 0!) . (3)7 (-2)0 + 10! / (7! 1! 2!). (3)1 (-2)2 + 10! /(5! 4! 1!). 34 (-2)1 (10. 9. 8 7!) / (7!.3.2.1).37 + (10. 9. 8. 7!) / (7! . 2) . 3 . 4 -[(10. 9. 8. 7. 6. 5! ) / (5!. 4.3.2.1.)] .3. 2. = 10 . 9 . 4 . 36 + 10 . 9 . 4 . 3 . 4 – 10 . 9 . 7 . 6 . 33 . 2 = 10 . 9 . 4 (36 + 12 – 7 34)= 360 (729 + 12 – 567)= 62640
  29. 29. Some tips on the solution of binomial – coefficients:(1) If the difference of the lower suffixes of binomial coefficients ineach term is same.For Ex.: C1 C3 + C2 C4 + C3 C5 + ------ etc.Then :Case -1 : If each term is positive, then(1+x)n = C0 + C1x + C2x2 + ------------ Cn xn ----------------- (i)Interchanging ‘1’ and ‘x’:(x+1)n = C0 xn + C1 xn-1 + C2 xn-2 + --------- + Cn -------------(ii)Then multiplying (i) and (ii), and equate the coefficient to suitablepower of ‘x’ on both sides.Case –II : If terms are alternately positive and negativeThen:(1-x)n = C0 – C1 x + C2 x2 - -------------- + (-1)n Cn xn --------------- (1)and (x+1)n = C0 xn + C1 xn-1 + C2 xn-2 + ---------- + Cn ------------- (2)The multiplying (1) and (2), and equate the coefficient of suitablepower of ‘x’ on both sides.Note : [ (Odd – number) / 2] = 8 (2) If the sum of the lower suffixes of binomial – coefficients ineach term is same.For Ex.: C0 Cn + C1 Cn-1 + C2. Cn-2 + ------- + Cn C0Then:Case – 1 : If each term is positive, then
  30. 30. (1+x)n = C0 + C1 x + C2 x2 - -------------- + Cn xn --------------- (1)and (1+x)n = C0 xn + C1 x + C2 x2 + ---------- + Cn xn------------- (2)Then multiplying (i) and (ii), and equate the coefficient of suitablepower of ‘x’ on both sides.Case –II : If terms are alternately positive and negative,Th (1+x)n = C0 + C1 x + C2 x2 - -------------- + Cn xn --------------- (1)and (1-x)n = C0 - C1 x + C2 x2 + ---------- + (-1)n Cn xn------------- (2)Then multiplying (i) and (ii) and equating the coefficient of suitablepower of ‘x’ on both side. PROBLEMS(1) Show that the middle – term in the expansion of (1+x)2n is 1. 3. 5 ------- (2n-1) / (n!) . 2n xn, ‘n’ being a positiveinteger.Sol.: The no. of terms in (1+x)2n = 2n+1 (odd). It’s ,middle-term = (2n + 1) / 2 = (n+1)th term. 2n Tn+1 = Cn x n = 2n! / (n! x n!). xn = 2n (2n-1) ------ 4.3.2.1 / (n! x n!). xn = [{(2n-1) (2n-3) ----- 3.1.} { 2n (2n-2) ------ 4.2.}] / n(n! n!). x = [{1. 3. 5. ---- (2n-1)} 2n {1.2 ---- n}] / (n! x n!) . xn = [{1.3.5----(2n-1)}. 2n] / (n! x n!). xn = Tn+1 = 1. 3. 5 – (2n-1) / (n!). 2n xn
  31. 31. **(2) Find the term independent of ‘x’ in the expansion of(i) (1+x+2x3) [(3 x2 / 2) – (1/3x)] 9(ii) [( x1/3 / 2) + x-1/5] 8Sol.: (i) (1+x+2x3) [(3/2)x2 - (1/3x)] 9 = (1+x+2x3) { [(3/2)x2] 9 - 9C1 [(3/2)x2 ] 8 1/3x + ---------- + + 9C6 [(3/2)x2] 3 (1/3x)6 - 9C7 [(3/2)x2] 2 (1/3x)7 ---} = (1+x+2x3) { [(3/2)x2 ] 9 – 9C1 (37 / 28)x15 + ---- + 9C6 (1 x 1 / 23 x 33) – 9C7 1/ (22 x 35) 1/ x3 + ---- }Term independent of ‘x’ : 9 C6 x 1/ (23 x 33) - 9 C7 2 / (22 x 35)= 9! / (6! x 3!) . 1/ (8 x 27) - 9!/(7! x 2!) . 1/ (2 x 243)= (9. 8. 7. 6!) / ( 6!. 3. 2. 1.) x 1/(8. 27) - (9. 8. 7!) / (7! . 2). 1/(2.243)= 7 / 18 - 2 / 27 = 17 / 54(ii) [(1 / 2) x1/3 + x-1/5] 8Sol.: General Term Tr+1 = nCr [(1/2) x1/3] n-r. (x-1/5) r n-r -r = nCr [(1/2) n-r] x 3 x 5 Here n = 8
  32. 32. = 8Cr (1/2) 8-r x (8-r)/3 -r/5 40 -8r Tr+1 = 8Cr (1/2) 8-r x 15 ---------------> (i)Putting (40 – 8r) / 15 = 0, we have r = 5 From (i), Term independent of ‘x’ : T6 = 8C5 (1/2) 8-5 = 8! / (5! X 3!) . 1 / 23 = (8. 7. 6. !5) / (5!. 3.2.1) . 1 / 8 = T6 = 7 **(3) Find the coefficient of ‘x’ in the expansion of (1-2x3 + x5) [1 +(1/x)]8Sol.: (1-2x3 + 3x5) [1 + (1/x)] 8= (1-2x3 + 3x5) [1 + 8C1 (1/x) + 8C2 (1/ x2) + 8C3 (1/ x3 ) + 8C4 (1/x4)+ 8C5 (1/ x5 )+ --- + 8C8 (1/ x8)coefficient of x = -2. 8C2 + 3 8C4 = -2. 8! / (2! x 6!) + 3. 8! / (4! x 4!) = -2. (8. 7) / 2 + 3 (8. 7. 6. 5.) / (4.3.2.1) = -56 + 210 = 154 **(4) Prove that the ratios of the coefficient of x10 in (1-x2)10 and theterm independent of ‘x’ in [x – (2/x)] 10 is 1 : 32.
  33. 33. Sol.: In (1-x)2 : Tr+1 = 10 Cr (-1)r (x2)r Putting r = 5 T6 = -10C5 x10 Coefficient of x10 = -10C5 10In [x – (2/x)] : Tr+1 = Cr (-1)r (x)10-r (2/x)r = (-1)r 10Cr. 2r. x10-2rPutting 10 – 2r = 0 r = 5So term independent of x : T6 = (-1)5 10C5. 25 Hence their ratio = (-10C5) : (-32. 10C5) = 1 : 32 **(6) If third term in the expansion of (x + x logx)5 is 10,00,000. Findthe value of ‘x’.Sol.: Putting log10x = z in the given expression : We have : ( x + xz)5 T3 = T2+1 = 5C2 (x)5-2 (xz)2 = 5C2 x3. x2z = 5! / (2! x 3!) x2z+3 = (5 x 4) / 2! x2z+3 = T3 = 10x2z+3
  34. 34. 10,00,000 = 10. x2z+3 Or x2z+3 = 105 (10z)2z+3 = 105 or 102z2+3z = 105 2z2 + 3z = 5 [Log10x = z] or 2z2 + 3z – 5 = 0 or (z-1) (2z+5) = 0 z = 1, - 5 / 2 or log10x = 1 or log10x = - 5/2 since x = 10 or 10-5/2**Question If in the expansion of (1+x)m (1-x)n, the coefficientsof ‘x’ and ‘x2’ are ‘3’ and ‘-6’ res. Find the value of ‘m’.Sol.: (1+x)m (1-x)n = [mC0 + mC1x + mC2x2 + ---- + m Cm xm] [nCo – nC1x + nC2x2 + ------- + (-1)n nCnxn]Coefficient of x = mC1 . nCo – mC0. nC1 = m! / ( 1! (m-1)!) x 1 – 1 x n! / (1! (n-1)!)
  35. 35. = m – n = 3 --------------- (i)Coefficient of x2 = -mC1 x nC1 +nC0 x mC2 + mC0 x nC2 = - m!/ (1! (m-1)!) n! / (1! (n-1)!)+ 1 m! / (2! m-2!) + 1 n! / (2! n-2!) = -mn + m (m-1) / 2 + n(n-1) / 2 = -6 or – 2mn + m(m-1) + n(n-1) = -12 or -2mn + m2 – m + n2 – n = 12 or (m-n)2 – (m+n) = -12From (i), putting the value of (m-n) : - 9 + (m + n) = 12 or m + n = 21 -----------> (ii)eqn (i) + eqn(ii) = 2m = 24 m = 12Q8. If the coefficients of (2r + 1)th term and (r + 2)th term in theexpansion of (1 + x)43 are equal, find ‘r’.Sol.: In (1 + x)43 : T2r+1 = 43 C2r. x2r 43 Coefficient = C2r And Tr+2 = 43Cr+1 xr+1 Coefficient = 43CrAccording to the questions: 43 43 C2r = Cr+1
  36. 36. 2r + r + 1 = 43 or 3r = 42 r = 14**Q9. If the coefficient of ‘4’th and ‘13’th terms in the expansionof [x2 + (1/x)] n be equal, then find the term which independent of‘x’.Sol.: T4 = T3+1 = nC3 (x2)n-3. 1/ x3 Coefficient = nC3 T13 = T12+1 = nC12 (x2)n-12 1 / x12 Coefficient = nC12According to the question: n C3 = nC12 n = 12 + 3 n = 15 Expansion = [x2 + (1/x)]15 15Now Tr+1 = Cr. (x2)15-r. 1/ xr 15 Tr+1 = Cr. x30-3r -------------> (i)
  37. 37. Putting : 30 – 3r = 0 r = 10From (i) T11 = 15C10 = 15!/(10! x 5!) = (15 x 14 x 13 x 12 x 11) /(5 x 4 x 3 x 2 x 1) = 3003.**Q10. In the expansion of (a – b)n, n ≤ 5, if the sum of the 5th and6th terms is zero. Find ( a / b) in terms of ‘n’.Sol.: T5 = T4+1 = nC4 (a)n-4 (-b)4 T5 = nC4 an-4 b4 T6 = T5+1 = nC5 (a)n-5 (-b)5 = -nC5 an-5 b5 T5 + T6 = 0 n C4 an-4 b4 – nC5 an-5 b5 = 0 or nC4 an-4 b4 = nC5 an-5 b5 or n!/(4! x n-4!) an-4 = n!/(5! x n-5!) an-5 b or an-4 / (n-4)(n-5!) = an-5 / 5(n-5!) b or an-4 / an-5 = b / 5 (n-4) or a(n-4)-(n-5) = (n – 4) / 5 .b or a = (n – 4)/5 . b or a/b = (n – 4) / 5
  38. 38. Q11. Find the coefficient of xr in the expansion of [x + (1/x)] n, ifit occurs.Sol.: General term : Tp+1 = nCp (x)n-p (1/x) p Tp+1 = nCp xn-2p ---------------> (i) Putting n-2p = r p = (n - r) / 2From: (i) T (n-r) / 2 +1 = nC(n-r) / 2 xr Coefficient of xr = nC (n – r) / 2 **Q12 Prove that the coefficient of the term independent of ‘y’ in theexpansion of [(y + 1)/( y 2/3 – y1/3 + 1) - (y – 1) / (y – y1/2)]10 is210.Sol.: We have (y + 1) / (y 2/3 – y1/3 + 1) Putting y = t3, we have = (t3 + 13) / (t2 – t + 1) = (t + 1) (t2 – t + 1) / (t2 – t +1) = t+1
  39. 39. (y + 1) / (y2/3 – y1/3 + 1) = y1/3 + 1and Putting y = a2 in (y – 1) / (y – y1/2 ) : = (a2 – 1) / (a2 – a) = (a+1) (a-1) / [a (a-1)] = (a + 1) / a = 1 + 1 / a (y – 1) / (y – y1/2) = 1 + 1 /vy (y + 1) / (y2/3 – y1/3 +1) - (y – 1) / (y – y1/2)]10 = [y1/3 + 1 – 1 – (1/ y1/2)] 10 = (y1/3 – y-1/2 )10In ( y1/3 – y-1/2)10, Tr+1 = 10Cr (y1/3)10-r. (-y-1/2)r = (-1)r 10Cr. (10-r) / 3 - r/2Tr+1 = (-1)r 10Cr. y(20-5r) / 6Putting (20 – 5r) / 6 = 0 or r=4Putting this value in (1) T5 = (-1)4 10 C4 = 10!/ (6! x 4!) = (10 x 9 x 8 x 7) /(4 x 3 x 2 x 1) T5 = 210
  40. 40. **Q13: x4-r occurs in the expansion of [x + (1/ x2)] 4n, prove that itscoefficients is: = (4n!) /[ (4/3)n-r]! x [(4/3)(2n+r)]!Sol.: In [x + (1/x2)]4n, Tp+1 = 4n Cp (x)4n-p (1/ x2)p 4n Tp+1 = Cp x4n-3p ----------> (i)Putting : 4n – 3p = 4r or 4 ( n-r ) / 3 =p 4nFrom (i) Tp+1 = C4(n-r) / 4. x4rCoefficient of x4r = 4n C4 (n-r) / 3 = (4n!) / [(4/3)n-r]! x [(4n/1) – 4(n-r)/3]! = (4n!) / [(4/3)n-r]! x [(4/3) 2(n+r)]!**Q14. Find the coefficient of x50 in (1+x)41 (1-x+x2)40.Sol.: (1+x)41 (1-x+x2)40 = (1+x) (1+x)40 (1-x+x2)40 = (1+x) [ (1+x) ( 1-x + x2)]40 = (1+x) (1+x3)40General Term = Tr+1 = (1+x) [40Cr (x3)r] = 40Cr (1+x) x3r = 40Cr (x3r + x3r + 1)
  41. 41. Here either 3r = 50 or 3r+1 = 50 r = (50 / 3) or (49 / 3)The value of ‘r’ is a fraction, so it doesn’t contain the term x 50. Socoefficient of x50 is ‘0’.Q15.: Show that that the term independent of ‘x’ in the expansion of 2n [x + (1/x)] is [1. 3. 5. ---- (2n-1) / (n!)] 2nSol.: General Term Tr+1 = 2nCr (x)2n-r (1/x) r = 2nCr. x2n-2r ---------> (i) Here 2n – 2r = 0 or n = rFrom (i) Tr+1 = 2nCn = 2n! / (n! x n!) = [2n (2n-1 ) ------ 3. 2. 1] / ( n! x n!) = { 2n (2n-2) ---- 4. 2 } { (2n-1) (2n-3) -----3.1.} / (n! x n!) = [2n {n (n-1) -----2.1.}] { (2n-1) -------4.3.1.} / (n! x n!) = 2n. n!{(2n-1) ---- 5. 3. 1. / (n! x n!) = {1. 3. 5. ----- (2n -1)} 2n / n!
  42. 42. Q16. The 3rd, 4th and 5th terms in the expansion of (x+a)n arerespectively ‘84’, ‘280’ and ‘560’, find the value of ‘x’, ‘a’ and ‘n’.Sol.: Tr+1 = nCr xn-r. ar Putting r = 2, 3 and 4 respectively T3 = nC2 xn-2. a2 = 84 ------------(i) T4 = nC3 xn-3 a3 = 280----------(ii)and T5 = nC4 xn-4 a4 = 560 --------(iii)eqn (i) eqn(iii) : [nC2 xn-2 a2] [nC4 xn-4 a4] = 84 560 = n!/[2! (n-2)!] n! / [4! (n-4)!] . x2n-6 a6 =84 560 or n (n-1) / 2 n(n-1) (n-2) (n-3) / 4! x x2n-6 a6 = 84 . 560------- (iv)Squaring of eqn (ii), we have : (nC3 x n-3 a3)2 = 2802 n C3 . nC3 . x2n-6 . a6 = 2802 = n! / [3! (n-3)!] n! / [3! (n-3)!] x2n-6 a6= 2802 or n (n-1)(n-2) / 6 n(n-1) (n-2) (n-3) / 3! x2n-6 a6 =280 280 ------- (v)eqn (v) ÷ eqn(iv) : n2 (n-1)2 (n-2)2 / (6 3!) 2 4! / [n2(n-1)2 (n-2)(n-3)] = (280 280) / (84 560)or 4 (n-2) / 3 (n-3) = 5/3
  43. 43. or 4n – 8 = 5n – 15n = 7Putting this value in (i), (ii) and (iii) : 7 C2 x5 a2 = 84 --------------- (vi) 7 C3 x4 a3 = 280 ----------------(vii) 7 C4 x3 a4 = 560-----------------(viii)egn (vii) ÷ egn(vi): (7C3 x4 a3) / (7C2 x5 a2) = 280 / 84 [7! / (3! . 4!)a] / [7! / (2! . 5!)x] = 10 / 3 or 7! / (3! . 4!) x (2! . 5!) / 7! . a / x = 10 /3 or 5/3 a/x= 10 / 3or a = 2xPutting this value in egn (vi): 7 C2. x5. 4x2 = 84or 7! / (2! x 5!)x7 = 21 (7 x 63) / 2 x7 = 21x7 = 1 x = 1
  44. 44. Putting this value in (ix) = a = 2Q17. Let ‘n’ be a positive integer. If the coefficients of second, thirdand fourth terms in (1+x)n are in arithmetic progression, then find thevalue of ‘n’.Sol: General Term : Tr+1 = nCr xr 2nd Term : T2 = nC1 x Coefficient = nC1 3rd Term : T3 = nC2 x2 Coefficient = nC2 Similarly coefficient of 4th term = nC3 These are in A. P., so. 2 nC2 = nC1 + nC3 2 [n! / {2! x (n-2)!}] = n! / {1! x (n-1!)} + n! / {3! x(n-3!)}or n! / (n-2!) = n! [1 / (n-1!) + 1 / {6 (n-3!}]or 1/ [(n-3) x (n-3!) = 1 / [(n-1)x (n-2) x(n – 3!)] +1/ [6! (n-3!)])or 1 / (n – 2) - 1 / [(n-1) (n-2)] = 1/6or (n – 1 – 1) / [(n-1) (n-2)] = 1/6
  45. 45. or (n- 2) / [(n-1) (n-2)] = 1/6or n–1 = 6 n = 7**Question Find the co-eff. Of x50 in the expansion of (1+x)1000+ x(1+x)999+ x2(1+x)998+........+x1000.SOL. It is in G.P. with ratio = x/(1+x) , n= 1001 , a= (1+x)1000 Sum = = (1+x)1001 – x1001 ,co-eff. Of x50 = 1001C50 .**Q18. The 6th term in the expansion of [(1/ x8/3) + x2 log10x]8 is5600. Prove that x =10. 8Sol.: T6 = T5+1 = C5 (1/ x8/3) 8-5 ( x2 log10x ) 5or 8C5 x (1 / x8) x c10 x (log10x)5 = 5600 8! / (5! x 3!) x c2 (log10x)5 = 5600 8. 7. 6. / 6 x c2 (log10x)5 = 5600or x2(log10x)5 = 100 = 102Clearly x = 10 satisfied as log1010 = 1.If x > 10 or < 10, the result will change in inequality.
  46. 46. Q1 What is the 10th term in the expansion of (x-1)11 (in decreasingpowers of x)?A-xB-11xC-x2D-55x2ANSWER (D)Hence, the 10th term in the expansion of(x-1)11 would be when k = 9Q2 What is the value of
  47. 47. (Dont worry if you get it wrong ... you can learn from your mistakes.)A36B-16C16DZeroANSWER (D)Both (1+i)6 and (1-i)6 each contain seven terms.The second, fourth, and sixth terms of (1-i)6 would be negative, andthese would cancel with the second, fourth and sixth terms of (1+i)6.Hence, the sum of the two expansions would bei is the square root of -1,
  48. 48. A8B7C-7D-8Q 4 Use the binomial theorem formula to determine the fourth term in theexpansion AnswerApplication of the Binomial TheoremFor situations involving distribution of a net charge over an extendedregion, the calculated electric field dependence may be checked inthe limit where the point of evaluation is far from the chargedistribution. When a finite amount of charge is the source, the far-field behavior of the electric field should behave as a point charge of
  49. 49. that amount. Often the binomial theorem is useful in consideringthat limit:The binomial theorem is especially useful in converting negative orfractional exponents into ordinary polynomial expressions from whichthe leading-order dependence may be determined. For example, forsmall x the binomial theorem can be used on the following fractionsto determine the linear dependence on x:
  50. 50. Multiple angle identitiesFor the complex numbers the binomial theorem can be combined withDe Moivres formula to yield multiple-angle formulas for the sine andcosine. According to De Moivres formula,Using the binomial theorem, the expression on the right can beexpanded, and then the real and imaginary parts can be taken to yieldformulas for cos(nx) and sin(nx). For example, since
  51. 51. De Moivres formula tells us thatwhich are the usual double-angle identities. Similarly, sinceDe Moivres formula yieldsIn general,andSeries for eThe number e is often defined by the formulaApplying the binomial theorem to this expression yields the usualinfinite series for e. In particular:The kth term of this sum is
  52. 52. As n → ∞, the rational expression on the right approaches one, andthereforeThis indicates that e can be written as a series:Indeed, since each term of the binomial expansion is an increasingfunction of n.We can write the binomial theorem as:Where n is a positive integer, and k is a nonnegative integer, 0, 1, ...,n and is the term number.If we let a=b=1, we find (1+1)n=2n is the sum of the terms, becausethe powers of a and b are all 1, and only the coefficients remain.Naturally, the values of the coefficients are not changed by thevalues of a and b, so the sum of the coefficients is always 2n,whatever the values of a and b.If we write a=1 and b=-1, then (1-1)n=0. We also notice that the evenpowers of b will be positive and the odd powers will be negative.

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