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Binomial theorem for any index for entrance exams.

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- 1. BINOMIAL THEOREM FOR ANY INDEX Let n be a rational number and x be a real number such that |x| < 1 , then (1+x)n = 1+ nx + x2 + . . . + xr + ... terms upto ∞ ......(1) Observations:o General term of the series (1+x)-n = Tr+1 = (-1)r xro General term of the series (1-x)-n = Tr+1 = xro If first term is not 1, then make first term unity in the following way: n n n (a+ x) = a (1+ ) if < | |< 1 IMPORTANT EXPANSIONS · (1+ x)-1 = 1- x +x2 –x3 + . . . + (-1)rxr+. . . · (1 - x)-1 = 1+ x +x2 +x3 + . . .+ xr + . . . · (1+ x)-2 = 1- 2x +3x2 –4x3+ . . .+ (-1)r(r+1)xr+. . . · (1 - x)-2 = 1+ 2x +3x2 +4x3 + . . .+ (r+1)xr+. . . · (1+x)-3 = 1- 3x +6x2 –10x3 +. . .+ (-1)r xr+. . . · (1-x)-3 = 1+ 3x +6x2 +10x3 + . . .+ xr+. . . · In general coefficient of xr in (1 – x)– n is n + r –1Cr . · (1 – x)–p/q = 1 + p/1! (x/q) + p(p+q)/2! (x/q)2 + ……. From (1) · (1 + x)–p/q = 1 – p/1! (x/q) + p(p+q)/2! (x/q)2 – ……. · (1 + x)p/q = 1 + p/1! (x/q) + p(p–q)/2! (x/q)2 + ……. · (1 – x)p/q = 1 – p/1! (x/q) + p(p–q)/2! (x/q)2 – ……. Solved Example 1: If –1 < x < 1, show that (1 –x)-2 = 1 + 2x + 3x2 + 4x3 + …..to ∞. Solution: We know that if n is a negative integer or fraction (1+x)n= 1 + Provided –1 < x < 1 Putting n = –2 and –x in place of x, we get (1+x)2 = 1 + = 1 + 2x + 3x2 + 4x3 + … to ∞. Properties Of Binomial Expansion 1. There are (n + 1) terms in the expansion of (a + b) n, the first and the last term being an and bn respectively. If nCx = nCy, then either x = y or x + y = n n Cr = nCn-r = n!/r!(n-r)! .(coeffts. Of terms equidistant from beginning and the end are equal) 2. The general term in the expansion of (a + x) n is (r + 1)th term given as
- 2. Tr+1 =nCran-r xr. Similarly the general term in the expansion of (x + a) n is given asTr+1 = nCrxn-r ar. The terms are considered from the beginning.3. The binomial coefficient in the expansion of (a + x)n which are equidistantfrom thebeginning and the end are equal i.e. nCr = nCn-r. n n n+1Note: Here we are using Cr + Cr-1 = Cr this concept will be discussed later in this chapter. m m+1Also, we have replace C0 by C0 because numerical value of both is same i.e. 1. Similarly we m m+1replace Cm by Cm+1.Properties of Binomial CoefficientsFor the sake of convenience, the coefficients nCo, nC1, ..., nCr, ..., nCn are usuallydenoted by Co, C1,..., Cr, ..., Cn respectivelyPut x = 1 in (A) and get, 2n = Co + C1 + ... + Cn ... (D)Also putting x = -1 in (A) we get,0 = Co - C1 + C2 - C3 + ......=> C0 + C2 + C4 + ...... + C1 + C3 + C5 +...... = 2n.Hence Co + C2 + C4 +...... = C1 + C3 + C5 +...... = 2n-1.n n-1 n-2 Cr = . Cr-1 = . Cr-2 and so on..... = , r = 1,2,3.....EXAMPLE: If C0 , C1 , C2 ... Cn denote the binomial coeffts. In the expansion of(1+x)n , prove that: +2 +3 +.........+n =(ii) (C0+ C1)( C1+ C2)( C2+ C3)..........( Cn-1+ Cn) =SOLUTION: use = , r = 1,2,3.....r. = n-r+1, r=1,2,3....n
- 3. = = n+(n-1)+.....+3+2+1(ii) ( ).........( ) = )(1+ ).........(1+ ) (either can take only L.H.S. &mult.And divide by c0.c1.c2.....cn ) now use above result.Illustration:If (1 + x)n = C0 + C1x + C2x2 +...+ Cnxn, then prove that C0 + (Co + C1) + (C0 +C1+ C2) + ... (C0 + C1 + C2 +...+ Cn-1) = n2n-1 (where n is an even integer.Solution: C0 + (C0 + C1) + (C0 + C1 + C2) +... (C0 + C1 + C2 +...+ Cn-1) = C0+(C0 + C1 + C2 +...+ Cn-1)+(C0 + C1)+(C0 + C1 + C2 +...+ Cn-2)+... = (C0 + C1 + C2 +...+ Cn)+ (C0 + C1 + C2 +...+ Cn)+... n/2 times = (n/2)2n = n . 2n-1Sum Of Binomial CoefficientsPutting x = 1 in the expansion (1+x) n = nC0 + nC1 x + nC2 x2 +...+ nCx xn, weget,2n = nC0 + nC1 x + nC2 +...+ nCn.We kept x = 1, and got the desired result i.e. ∑ nr=0 Cr = 2n. Note: This one is very simple illustration of how we put some value of xand get the solution of the problem. It is very important how judiciously youexploit this property of binomial expansion.Illustration: Find the value of C0 + C2 + C4 +... in the expansion of (1+x)n.Solution:We have, (1 + x)n = nC0 + nC1 x + nC2 x2 +... nCn xn.Now put x = -x; (1-x)n = nC0 - nC1 x + nC2 x2 -...+ (-1)n nCn xn.
- 4. Now, adding both expansions, we get, (1 + x)n + (1-x)n = 2[nC0 + nC2 x2 + nC4 x4 +.........]Put x=1 => (2n+0)/2 = C0 + C2 + C4 +...... or C0 + C2 + C4 +......= 2n-1We have found the sum of binomial coefficients. But if these coefficients aremultiplied by some factors can we find the sum for such expressions? Yes, we can often find it by creatively applying what we have learnt. Letus see how differentiation and integration are useful in these situations. Suppose we want to calculate the value of ∑nr=0 Cr i.e. 0.C0 + 1 C1 + 2C2 +...+ n nCn If we take a close look to the sum to be found, we find that coefficientsare multiplied with respective powers of x.If we want to multiply the coefficient of x by its power differentiation is ofhelp. Hence differentiate both sides of (1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 +...+ nCn xn, with respect to x weget n(1 + x)n-1 = 1 C1 x1-1 + 2 C2 x2-1 +...+ n Cn xn-1 Put x = 1, we get, n 2n-1 = + 1 C1 + 2 C2 +...+ n Cn.Or, ∑nr=0 r Cr = n 2n-1, which is the answer. Important: It has said earlier in this chapter that we should use andexploit the property that x can take any value in the expansion of (1 + x)n. Now, let us try to find the value of C0 - C2 + C4-C6 +......... Analysing the above, expression, we find that C 0, C2, C4 are allcoefficients of even powers of x. Had it been like C0 + C2 + C4 +...... we could have evaluated simply by themethod described earlier or had it been like C0-C1 + C2-C3 +... we could haveput x = -1 in the expansion of (1 + x)n and find the sum.
- 5. But here the case is different. The expression consists of coefficients of onlyeven powers. In such cases when there is alternative sign charge for thecoefficient, of same nature of powers (even or odd) use of I (iota) comes toour rescue. (1 + x)n = C0 + C1 x + C2 x2 +...... (1 - x)n = C0 - C1 x + C2 x2 +......By adding (1 + x)n + (1 - x)n = 2[C0 + C2 x2 + C4 x4 +......]. Clearly use of √-1 (iota) will generate the given expression. ((1+i)n + (1-i)n)/2 ] [C0 - C2 + C4 + ......]1. Divisibility problems: Let (1 + x)n = 1 + nC1x + nC2x2 +...+ nCnxn. In any divisibility problem, we have to identify x and n. The number bywhich division is to be made can be, x, x2 or x3, but the number in the base isalways expressed in form of (1+x).Illustration: Find the remainder when 7103 is divided by 24.Solution: 7103 = 7(50 - 1)51 = 7(5051 - 51 C1 5050 + 51 C2 5049 - ... - 1) = 7(5051 - 51 C15050 +...+ 51 C5050) - 7 - 18 + 18 = 7(5051 - 51 C15050 +...+ 51 C5050) - 25 + 18 => remainder is 18.Multinomial Expression If such a case arises, then it is not called Binomial Expansion, it is calledMultinomial Expansion. If n ε N, then the general term of the multinomialexpansion (x1 + x2 + x3 +...+ xk)n is (n!/a1!a2!...ak!) (x1a1,x2a2,..…xkak ), wherea1 + a2 + a3+......+ ak = n and a < ai < n, I = 1, 2, 3, ...... k and the totalnumber of terms in the expansion is n+k-1Cn-1.
- 6. Problem Related to Series of Binomial Coefficients in Which Each Term is aProduct of two Binomial Coefficients.(a) If sum of lower suffices of binomial expansion in each term is the same i.e. nC0 nCn + nC1 nCn-1 + nCn-2 +...+ nCn nC0 i.e. 0 + n = 1 + (n-1) = 2 + (n-2) = n + 0. Then the series represents the coefficients of xn in the multiplication ofthe following two series (1+x)n = C0 + C1x + C2x2 +...+ Cnxn and (1+x)n = C0 + C1x + C2x2 +...+ Cnxn.Illustration:Prove that C0Cr +C1Cr+1 +C2Cr+2 +...+Cn-r Cn =2n!n-r!n+r!(Similarily we can prove that C0C1 + C1C2 + C2C3 +...+ Cn-1 Cn =Solution: We have, C0 + C1x + C2x2 + ... + Cnxn = (1+x)n ... (1) Also C0xn + C2xn-2 +...+ Cn = (x+1)n ... (2) Multiplying (1) and (2), we get (C0 + C1x2 +...+ Cnxn)(C0xn + C1xn-2 + C2xn-2 +...+ Cn) = (1+x)2n ... (3) Equating coefficient of xn-r from both sides of (3), we get C0Cr + C1Cr+1 + C2Cr+2 +...+ Cn-rCn = =Illustration:
- 7. (i) Prove that Co2 + C12 +...+ Cn2 = (2n)!/n!n! . (ii) prove that Co2 - C12 + ...+(-1)n Cn2 = if n is evenSolution: (i) Since (1 + x)n = C0 + C1x + C2x2 +...+ Cnxn, ... (1) (x + 1)n = C0xn + C1xn-2 +...+ Cn, ... (2) (C0 + C1x + C2x2 +...+ Cnxn)(C0xn + C1xn-1 + C2xn-2 +...+ Cn)=(1+x)2n. Equating coefficient of xn, we get C02 + C12 + C22 +...+ Cn2 = 2n Cn = (2n)!/n!n! . (ii) (x+1)n(1-x)n = (C0xn + C1x(n-1) + C2xn-2 +...+ Cn-1x+Cn )( C0 - C1x + C2x2 -C3x3+C4x4...+(-1)n Cnxn )= (1-x2)n = C0 - C1x2 + C2x4 -C3x6+...+(-1)n Cnx2nEquating the coeffts of xn on both sidesCo2 - C12 + ...+(-1)n Cn2 = 0, if n is oddIf n is even , suppose (r+1)th term in the expansion of (1-x2)n contains xn, we have Tr+1 = (-1)r x2r it implies 2r =n i.e., r=n/2 hence proved.Example: Find the coeffts. Of x 4 in the expansion of (1+x)n(1-x)n.Deduce that C2 = C0C4 - C1C3 + C2C2 – C3 C1+ C4C0SOLUTION: (1+x)n(1-x)n = (C0 + C1x + C2x2 +C3x3+C4x4...+ Cnxn )( C0 - C1x +C2x2 -C3x3+C4x4...+(-1)n Cnxn )Coeffts of x4 on R.H.S. = C0C4-C1C3+C2C2-C3C1+C4C0(1-x2)n = C0 - C1x2 + C2x4 -C3x6+...+(-1)n Cnx2nIllustration: If (1+x)n = , then prove that m Cr nC0 + m Cr-1 nC1 + m Cr-2 nC2 +...+ m C1 nCr-1 + m C0 nCr = m+n Cr where m, n, r are positive integers and r < m and r < n.
- 8. Solution: (1+x)n = nC0 + nC1x + nC2x2 +...+ nCrxr +...+ nCnxn ... (1) and also (1+x)m = m C0 + m C1x + m C2x2 +...+ m Crxr +...+ m Cmxm ... (2) Multiplying (1) and (2), we get (nC0 + nC1x + nC2x2 +...+ nCrxr +...+ nCnxn)x (mC0 + m C1x + m C2x2 +...+ m Crxr +...+ m Cmxm) = (1+x)m+n m+n m+2 m+n = C0 + C1x + C2x2 +...+ m+n Crxr +...+ m+n Cm+nxm+n Equating the coefficient of xr, we get m Cr nC0 + m Cr-1 nC1 + m Cr-2 nC2 +...+ m C1 nCr-1 + m C0 nCr = m+n Cr(b) If one series has constant lower suffices and other has varying lowersufficesIllustration: Prove that nC0.2nCn - nC12n-2Cn + nC2.2n-4Cn -...= 2n.Solution: n C0.2nCn - nC12n-2Cn + nC2/2n-4Cn -... = coefficient of xn in[nC0(1+x)2n - nC1(1+x)2n-2+nC2(1+x)2n-4 - ...] = coefficient of xn in [nC0((1+x)2)n - nC1((1+x)2)n-1 + nC2((1+x)2)n-2 - ...] = coefficient of xn in [(1+x)2 - 1]n = coefficient of xn in (2x+x2)n = co-efficient of xn in xn (2+x)n = 2n.Some Important Results
- 9. 1. Differentiating (1+x)n = C0 + C1x + C2x2 +...+ Cnxn of both sides wehave, n(1 + x)n-1 = C1 + 2C2x + 3C3x2 +...+ nCnxn-1. ... (E) Put x = 1 in (E) so that n2n+1 = C1 + 2C2 + 3C3 + ...+ nCn. Put x = -1 in (E) so that 0 = C1 - 2C2 +...+ (-1)n-1 nCn. Differentiating (E) again and again we will have different results.2. Integrating (1 + x)n, we have, ((1+x)n+1)/(n+1) + C = C0x + C1x2/2 + C2x3/3 +.....+Cnxn+1/(n+1) (where C is a constant) For x = 0, we get C = -1/(n+1) . Therefore ((1+x)n+1 - 1)/(n+1) = C0x + C1x2/2 + C2x3/3+.....+Cnxn+1/(n+1) ..... (F) Put x = 1 in (F) and get (2n+1 - 1)/n+1= C0 + C1/2 +...Cn/n+1 . Put x = -1 in (F) and get, 1/n+1 = C0 - C1/2 + C2/3 - ...... Put x = 2 in (F) and get, (3n+1-1)/n+1 = 2 C0 + 22/2 C1 + 22/3C2 +...+ 2n+1/n+1 = Cn.Problems Related to Series of Binomial Coefficients in Which Each Termis a Product of an Integer and a Binomial Coefficient, i.e. In the Formk.nCr.Illustration: n If (1+x) = xr then prove that C1 + 2C2 + 3C3 +...+ nCn = n2n-1.Solution: Method (i) : rth term of the given series rth term of the given series, tr = nCr n-1 n-1 tr = r × n/r × Cr-1 = n × Cr-1 (because nCr =n/r .n-1Cr-1) Sum of the series =
- 10. Put x = 1 in the expansion of (1 + x) n-1, so that (n-1C0 + n-1 C1 +...+ n-1 Cn-1) = 2n-1 => = n.2n-1. Method (ii) : By Calculus We have (1 + x)n = C0 + C1x + C2x2 +...+ Cnxn. ... (1) Differentiating (1) w.r.t. x, we get n(1 + x)n-1 = C1 + 2C2x + 3C3 x2 +...+ n Cnxn-1. ... (2) Putting x = 1 in (2), we have, n 2n-1 = C1 + 2C2 +....+ nCn. ... (3)Illustration:If (1+ x)n = xr then prove that C0 + 2.C1 + 3.C2+...+(n+1)Cn=2n-1(n+2).Solution: Method (i): rth term of the given series rth term of the given series tr = nCr-1 = [(r-1) + 1]. nCr-1 = (r-1) nCr-1 + nCr-1 = n. n-1 Cr-2 + nCr-1 (because nCr-1 =n/(r-1) . n-1 Cr-2) Sum of the series = = n[n-1C0 + n-1 C1 +...+ n-1 Cn-1]+[nC0 + nC1 +...+ nCn] = n.2n-1 + 2n = 2n-1 (n+2). Method (ii) by Calculus. We have (1 + x)n = C0 + C1x + C2x2 +...+ Cnxn. ... (1) Multiplying (1) with x, we get x(1+x)n = C0x + C1x2 + C2x3 +...+ Cnxn+1. ... (2) Differentiating (2) w.r.t. x, we have (1 + x)n + n(1 + x)n-1 x = C0x + 2C1x2 +...+ (n+1)Cnxn ...(3)
- 11. Putting x = 1 in (3), we get 2n + n.2n-1 = C0 + 2C1 + 3C2 +...+ (n+1)Cn => C0 + 2C1 + 3C2 +...+ (n+1)Cn = 2n-1 (n+2).Problems Related to Series of Binomial Coefficient in Which Each Term isBinomial Coefficient divided by an Integer, i.e. in the Form of nCr/k.Now, think how to find the following sum C0/1 + C1/2 + C2/3 +......+ Cn/n+1 = ? In this sum coefficients are divided by the respective power of x +1. This expression can be achieved by Integrating the expansion of (1 +x)n under proper limits. = = = [ .....{ use = . , and add & subtract }= [2(n+1) -1]Similarily we can prove C0/1 - C1/2 - C2/3 +......+(-1)n Cn/n+1 =1/(n+1)L.H.S. = = =[ ....Or another methodIllustration: If (1+x)n =Solution:
- 12. Method (i) : rth term of the given series Method (ii): By Calculus (1+x)n = C0 + C1x + C2x2 +...+ Cnxn ... (1) Integrating both the sides of (1) w.r.t. x between the limits 0 to x,we get ... (2) Substituting x = 1 in (2), we get 2n+1/n+1 = C0 + c1/2 + c2/3+.....+ cn/n+1 .Illustration: If (1+x)n = xr ,show thatMethod I : rth term of the given series Tr =
- 13. Method II : (By Calculus) (1 + x)n = C0 + C1x + C2x2 +...+ Cnxn => x(1+x)n = C0x + C1x2 + C2x3 +...+ Cnxn+1 ... (1) Integrating both the sides of (1) with respect to x Put x = 0, => k = 1/((n+1)(n+2)) Put x = 1, .Greatest Binomial CoefficientTo determine the greatest coefficient in the binomial expansion, (1+x) n,when n is a positive integer. Coefficient of(Tr+1/Tr) = Cr/Cr-1 = (n-r+1)/r = ((n+1)/r) - 1.Now the (r+1)th binomial coefficient will be greater than the rthbinomial coefficient when, Tr+1 > Tr=> ((n+1)/r)-1 > 1 => (n+1)/2 >r. ....... (1)But r must be an integer, and therefore when n is even, the greatestbinomial coefficient is given by the greatest value of r, consistent with(1) i.e., r = n/2 and hence the greatest binomial coefficient is nCn/2.Similarly in n be odd, the greatest binomial coefficient is given when,r = (n-1)/2 or (n+1)/2 and the coefficient itself will be nC(n+1)/2 or nC(n-1)/2, both being are equal
- 14. Note: The greatest binomial coefficient is the binomial coefficient ofthe middle term. Illustration: Show that the greatest the coefficient in the expansion of (x +1/x)2n is (1.3.5...(2n-1).2n)/n! .Solution: Since middle term has the greatest coefficient. So, greatest coefficient = coefficient of middle term 2n = Cn = (1.2.3...2n)/n!n! = (1.3.5...(2n-1).2n)/n!.Numerically greatest termTo determine the numerically greatest term in the expansion of (a + x) n,where n is a positive integer.ConsiderThus Note : {((n+1)/r) - 1} must be positive since n > r. Thus Tr+1 will bethe greatest term if, r has the greatest value as per the equation (1).Illustration: Find the greatest term in the expansion of (3-2x)9 when x = 1.Solution:
- 15. Tr+1/Tr = ((9-r+1)/r) . (2x/3) >1 i.e. 20 > 5r If r = 4, then Tr+1 = Tr and these are the greatest terms. Thus4th and 5thterms are numerically equal and greater than any other termand their value is equal 39 × 9C3 × (2/3)3 = 489888.Illustration: Find the greatest term in the expansion of (2 + 3x) 9 if x = 3/2.Solution: Here Tr+1/Tr = ((n-r+1)/r)(3x/2) = ((10-r)/r)(3x/2), (where x =3/2) = ((10-r)/r)(3x/2)(3/2) = ((10-r)/r).9/4 = (90-9r)/4r Therefore Tr+1 > Tr, if 90 - 9r > 4r. => 90 > 13r => r < 90/13 and r being an integer, r = 6. Hence Tr+1 = T7 = T6+1 = 9C6 (2)3 (3x)6 = 313.7/2.Illustration:Given that the 4th term in the expansion of (2 + (3/8)x)10 has themaximum numerical value, find the range of values of x for which thiswill be true.Solution: Given 4th term in (2 + (3/8)x)10 = 210 (1 + (3/16)x)10, isnumerically greatest =>|T4/T3| > 1 and |T5/T4| < 1 => =>|x| > 2 and |x| < 64/21 => x ε [-(64/21),-2]U[2,(64/21)].Illustration:
- 16. Find the greatest term in the expansion of √3(1 + (1/√3)) 20.Solution: Let rth term be the greatest Since Tr/Tr+1 = (r/(21-r)).√3.Now Tr/Tr+1 > 1 => r > 21/(√3+1) ...... (1)Now again Tr/Tr+1 = (r/(21-r)) √3 < 1 => r < (22+√3)/(√3+1)...... (2)from (1) and (2)22/(√3+1) < r < (22+√3)/(√3+1) 20=> r = 8 is the greatest term and its value is √3 . C7(1/√3)7 = 20 C7 (1/27).Illustration:Find the first negative term in the expansion of (1+x) 7/2.Solution: Tr+1 = this will be first negative term if <0 it implies r>9/2 or r≥5 , hence 6th term is the firstnegative term = -7/256 x5 ( by putting r=5).Illustration:If (1-x)-n = a0+a1x+a2x2+a3x3+......., find the value ofa0+a1+a2+......+an.Solution: use (1-x)-n(1-x)-1 =(a0+a1x+a2x2+.....+anxn+...)(1+x+x2....+xn+....)Coeffts. Of xn in (1-x)-(n+1) = a0+a1+a2+......+anTr+1==Put r=n then coeffts. Of xn in (1-x)-(n+1)= a0+a1+a2+......+an= = (2n)!/(n!)2Illustration:
- 17. Find the coefft. Of x6 in (1+x+x2)-3Solution: (1+x+x2)-3 ={ }-3 = =(1-x)3(1-x3)-3Coefft. Of x6 is 6-3=3.Particular CasesWe have (a + x)n = an + nC1 an-1 x + nC2 an-2 xr +...+ xn. ...... (1)(i) Putting x = -x in (1), we get (a-x)n = an - nC1 an-1x + nC2 an-2 x2 - nC2 an-3 x3 +...+ (-1)r nCr an-r x +...+ (-1)n xn. r(ii) Putting a = 1 in (1), we get, (1+x)n = nC0 + nC1x + nC2x2 +...+ nCr xr +...+ nCnxn. ... (A)(iii) Putting a = 1, x = -x in (1), we get (1-x)n=nC0-nC1 x + nC2 x2-nC3x3 +... (-1)r nCr xr +... (-1)n nCnxn ...(B)Tips to Remember(a) Tr/Tr+1 = ((n-r+1)/r ).(x/a) for the binomial expansion of (a + x)n. (n+1)(b) Cr = nCr + nCr-1.(c) r nCr = n n-1 Cr-1(d)(e) When n is even, (x + a)n + (x - a)n = 2(xn + nC2 xn-2 a2 + nC4 xn-4 a4 +...+ nCn an). When n is odd, (x + a)n + (x - a)n = 2(xn + nC2 xn-2 a2 +...+ nCn-1 x an-1). When n is even (x + a)n - (x - a)n = 2(nC1 xn-1 a + nC3 xn-3 a3 +...+ nCn-1 x an-1).
- 18. When n is odd (x + a)n - (x - a)n = 2(nC1 xn-1 a + nC3 xn-3 a3 +...+ nCn an).Binomial Theorem(Approximation)Illustration:If p is nearly to q and n>1, show that = ,hence find .Solution: let p = q+h, where h is very small that its square & higherpowers may be neglected. Then ,L.H.S. = = = -1={ }={1+h/(nq)} [expandingneglecting higher power of h]R.H.S. = = = 1+h/(nq) [ same reason]Put p=99, q= 101 and n=6 in above result , we get =599/601. Illustration:Assuming x to be so small that x2 and higher power of x can beneglected, show that: is approx. Equal to 1- (305x/96).Solution: == =(1-3x)(1-3x/32)(1-x/12) [by nusing (1+x) =1+nx] =(1-3x-3x/32)(1-x/12)=(1-99x/32-x/12)=1-305x/96. [by neglecting higher power of x]Similarily, we can find a and b , if = a+bx for allsmall values of x , a= 1, b= -305/96 by equating first two terms
- 19. on the both sides.SOME EXAMPLESExample1 : If a1, a2, a3 and a4 are the coefficients of any four consecutiveterms in the expansion of (1+x)n then prove that: a1/(a1+a2) + a2/(a3+a4) = 2a2/(a2+a3)Solution: As a1, a2, a3 and a4 are coefficients of consecutive terms, then Let a1 = nCr a2 = nCr+1 a3 = nCr+2 and a4 = nCr+3 Now a1/(a1+a2) = nCr/(nCr+nCr+1) = 1/(1+((n-r)/(r+1))) =(r+1)/(n+1) Similarly, a2/(a2+a3) = (r+3)/(n+1) Now a3/(a3+a4) + a1/(a1+a2) = (2r+4)/(n+1) = 2(r+1)/(n+1) = 2a2/(a2+a3) (Hence,proved)Example2 : Find out which one is larger 9950 + 10050 or 10150.Solution: Lets try to find out 10150 - 9950 in terms of remaining termi.e. 10150 - 9950 = (100+1)50 - (100 - 1)50 = (C0.10050 + C110049 + C2.10048 +......) = (C010050 - C110049 + C210048 -......) = 2[C1.10049 + C310047 +.........]
- 20. = 2[50.10049 + C310047 +.........] = 10050 + 2[C310047 +............] > 10050 => 10150 > 9950 + 10050Example3 : Find the value of the greatest term in the expansion of√3(1+(1/√3))20.Solution: Let Tr+1 be the greatest term, then Tr < Tr+1 > Tr+2 Consider : Tr+1 > Tr 20 => Cr (1/√3)r > 20Cr-1(1/√3)r-1 => ((20)!/(20-r)!r!) (1/(√3)r) > ((20)!/(21-r)!(r-1)!) r-1(1/(√3) ) => r < 21/(√3+1) => r < 7.686 ......... (i) Similarly, considering Tr+1 > Tr+2 => r > 6.69 .......... (ii) From (i) and (ii), we get r=7 Hence greatest term = T8 = 25840/9Example 4:Find the coefficient of x50 in the expansion of (1+x)1000 +2x(1+x)999 + 3x2(1+x)998 +...+ 1001x1000.Solution: Let S = (1 + x)1000 + 2x(1+x)999 +...+ 1000x999 (1+x) + 1001 1000x
- 21. This is an Arithmetic Geometric Series with r = x/(1+x) and d= 1. Now (x/(1+x)) S = x(1 + x)999 + 2x2 (1 + x)998 +...+1000x1000 + 1000x1001/(1+x) Subtracting we get, (1 - (x/(x+1))) S =(1+x)1000 + x(1+x)999 +...+ x1000 -1001x1000/(1+x) or S = (1+x)1001 + x(1+x)1000 + x2(1+x)999 +...+ x1000 (1+x)-1001x1001 This is G.P. and sum is S = (1+x)1002 - x1002 - 1002x1001 So the coeff. of x50 is = 1002 C50Example 5: n Show that Ck (sin kx) cos (n-k)x = 2n-1 sin(nx)Solution: n We have Ck sin kx cos (n-k)x n =1/2 Ck [sin (k x + nx - kx) + sin (kx - nx + kx)] n n =1/2 Ck sin n x + 1/2 Ck sin (2kx - nx) n = 1/2 sin n x Ck 1/2 [nC0 sin (-nx) + nC1 sin (2-n)x +... ...+ nCn-1 sin (n-2)x + nCn sin nx]= 2n-1 sin nx + 0 (as terms in bracket, which are equidistant, fromend and beginning will cancel each other.Example 6: If (15+6√6)2n+1 = P, then prove that P(1 - F) = 92n+1 (where Fis the fractional part of P).Solution: We can write
- 22. P = (15+6√6)2n+1 = I + F (Where I is integral and F is thefractional part of P) Let F = (15+6√6)2n+1 Note: 6√6 = 14.69 => 0 < 156√6 < 1 => 0 < (15+6√6)2n+1 < 1 => 0 < F < 1 Now, I + F = C0 (15)2n+1 + C1(15)2n 6√6 + C2 (15)2n (6√6)2 +... F = C0 (15)2n+1 - C1(15)2n 6√6 + C2(15)2n-1 (6√6 )2 +... I + F + F = 2[C0 (15)2n+1 + C2 (15)2n-1 (6√6 )2 +...] Term on R.H.S. is an even integer. => I + F + F = Even integer => F + F = Integer But, 0 < F < 1 and (F is fraction part) 0 < F < 1 => 0 < F + F < 2 Hence F + F = 1 F = (1-f) .·. P(1-F) = (15 + 6√6 )2n+1 (15-6√6 )2n+1 = (9)2n+1 Example 7: Using ∫01(tx+a-x)n dx,prove that ∫01xk (1-x)(n-k) n (-1) dx=[ Ck (n+1)]Solution:
- 23. The given integral can easily be evaluated, as follows: I = ∫01(tx+a-x)n dx = ∫01((t-1)x+1)n dx = [((t-1)x+1)(n+1)/((n+1)(t-1))]01 =(t(n+1)-1)/(n+1)(t-1)=1/(1+n) [1 + t + t2 +...+ tk +...+ nt ] ...... (i) Also, I = ∫01(tx+(1-x))n dx =∫01 n Ck (1-x)(n-K) tk xk dx ......(ii) Comparing the coefficient of tk from (i) and (ii), we get, ∫01 n Ck. xk (1-x)n-k dx=1/(n+1) => ∫01xk (1-x)n-k dx = 1/( nCk (n+1)) (Hence, proved)Example 8:Prove that (-3)r-1 3n C2r-1= where k = 3n/2 and n is an evenpositive integer.Solution: Since n is even integer, let n = 2m, k = 3n/2 = 6m/2 =3m The summation becomes, S= Now, (1+x)6m = 6m C0 + 6m C1 x + 6m C2 x2 +... 6m ...+ C6m-1 x6m-1 + 6m C6m x6m ...... (i) (1-x)6m = 6m C0 + 6m C1 (-x) + 6m C2 (-x)2 +... 6m ...+ C6m-1 (-x)6m-1 + 6m C6m (-x)6m ...... (ii)
- 24. => (1+x)6m - (1-x)6m = 2[6mC1x + 6m C3x2 +...+ 6m C6m-1x6m-2 ] ((1+x)6m - (1-x)6m)/2x = 6m C1 + 6m C3 x2 +......+ 6m C6m-1 6mx -2 Let x2 = y =>((1 + √y)6m-(1-√y)6m)/2√y=6mC1+6mC3y 6m 6m 3m-1+ C3 y+...... C6m-1 y With y = -3, RHS becomes = S. LHS =Example9: If (1+x)n = C0 + C1 x + C2 x2 +...+ Cn xn then show that (i) ∑0≤i<j≤n Ci Cj = 22n-1 - (ii) ∑0≤i<j≤n (Ci + Cj)2 = (n - 1) 2n Cn + 22n (iii) ∑0≤i<j≤n ∑Ci Cj = n (22n-1 - 1/2 2n Cn)Solution:(i) We have (C0 + C1 + C2 +...+ Cn)2 = C02 + C12 +...+ Cn2 + 2 ∑0≤i<j≤n ∑Ci Cj we get, (2n)2 = 2n Cn + 2 ∑0≤i<j≤n ∑Ci Cj therefore ∑0≤i<j≤n ∑Ci Cj = 22n-1 - (2n)!/2(n!)2(Hence, proved)(ii) ∑0≤i<j≤n ∑Ci Cj n(C02 + C12 + C22 +...+ Cn2) + 2∑0≤i<j≤n ∑Ci Cj
- 25. 2n = n Cn + 2{22n-1 - (2n)!/2(n!)2}[from part (i)] 2n =n Cn + 22n - 2n Cn 2n = (n-1) Cn + 22n(Hence, proved)(iii) Let ∑0≤i<j≤n ∑Ci Cj replace i by (n - i) and j by (n - j), we have P = ∑0≤i<j≤n ∑(2n-i0j) Cn-i Cn-j = ∑0≤i<j≤n ∑(2n-(i+j)) Ci Cj [·.· nCn = nCn-r] = 2n ∑0≤i<j≤n ∑Ci Cj - P .·. ∑0≤i<j≤n ∑(i+j) Ci Cj = n[22n-1 - (2n)!/2(n!)2]Example 10. Find the value of (0.98)-3 upto 2 decimal places.Solution: (1-0.02)-3= [1+(-3)(-0.02)+(-3)(-3-1)/2!(-0.02)2+......] =[1+0.0600+0.0024] 1.0624 = 1.06 { by neglectinghigher power ofmore zeros just after decimal}. By: --indu thakur

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