Ensuring Technical Readiness For Copilot in Microsoft 365
Binomial theorem
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. We have enough now to start talking about the pattern.
The Pattern
In the last result we got:
a3 + 3a2b + 3ab2 + b3
Now, 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. = 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).
Let's 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
wasn't shown):
4. They actually make Pascal's Triangle!
Each number is just the two numbers
above it added together (except for the
edges, 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
Pascal's 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. Permutations
The "!" means "factorial", for example 4! =
1×2×3×4 = 24
Example: Row 4, term 2 in Pascal's Triangle is "6". Let's 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. OK ... it won't make much sense without an example.
So let's try using it for n = 3 :
BUT ... it is usually much easier just to remember the patterns:
The first term's exponents start at n and go down
The second term's exponents start at 0 and go up
Coefficients are from Pascal's 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 1x054
Then 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. The exponents for x3 are:
(2x)345
The coefficient is "8 choose 3". We can use Pascal's 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, isn't it?
And one last, most amazing example:
Example: A formula for e (Euler's Number)
You can use the Binomial Theorem to calculate e (Euler's 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 let's use the Binomial Theorem:
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...
10. 1.)
Keep going, always
adding pairs of numbers
from the previous row..
To find, say, 6C4, you go
down to the row where
there is a "6" after the
initial "1", and then go
over to the 5th (not
the 4th) entry, to find that
6C4 = 15.
As you might imagine, drawing Pascal's Triangle every time
you have to expand a binomial would be a rather long process,
especially if the binomial has a large exponent on it. People
have done a lot of studies on Pascal's Triangle, but in practical
terms, it's probably best to just use your calculator to find nCr,
rather than using the Triangle. The Triangle is cute, I suppose,
but it's not terribly helpful in this context, being more time-
consuming than anything else. For instance, on a test, do you
want to evaluate "10C7" by calculating eleven rows of the
Triangle, 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 the
powers on each term in the expansion always added up to
whatever n was, and that the terms counted up from zero to n.
Returning to our intial example of (3x – 2)10, the powers on
every term of the expansion will add up to 10, and the powers
on the terms will increment by counting up from zero to 10:
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)10
Note how the highlighted counter number counts up from zero
to 10, with the factors on the ends of each term having the
counter number, and the factor in the middle having the counter
number subtracted from 10. This pattern is all you really need to
know 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 isn't 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
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)10
So the fourth term is not the one where I've counted up to
4, but the one where I've counted up just to 3. (This is
because, just as with Java script, the counting starts with
0, not 1.)
Note that, in any expansion, there is one more term than
the 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 terms
will count up, not from 1 to 10 or from 1 to 11, but from 0 to
10. This is why the fourth term will not the one where I'm
using "4" as my counter, but will be the one where I'm
using "3".
10
C3 (3x)10–3(–2)3 = (120)(2187)(x7)(–8) = –2099520x7
Find the tenth term in the expansion of (x + 3)12.
To find the tenth term, I plug x, 3, and 12 into the Binomial
Theorem, using the number 10 – 1 = 9 as my counter:
12
C9 (x)12–9(3)9 = (220)x3(19683) = 4330260x3
Find the middle term in the expansion of (4x – y)8.
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 I'll 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) = 17920x4y4
You 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, I'm 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)4
Don't let the Binomial Theorem scare you. It's just another
formula to memorize. A really complicated and annoying
15. formula, I'll grant you, but just a formula, nonetheless. Don't
overthink the Theorem; there is nothing deep or meaningful
here. Just memorize it, and move on
Binomial Theorem for positive integral index:
(x+y)n = xn + nC1xn-1y+nC2xn-2y2+-----+nCrxn-ryr+ -------+---------+nCn-
n-1
1xy + ncnyn.
It can be represented as:
(x+y)n = n
Crxn-ryr
Particular – 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. 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 equal
Q2: 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)2n
Sol.: (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)
18. (x+y)n is called general – term, and
it is denoted by
Tr+1 = nCrxn-ryr
Explanation: We know
(x-y)n = nCoxnyo+nC1xn-1y1+nC2xn-2y2+----+nCnxoyo
Here:
First term T1 = nCoxnyo
T2 = nC1xn-1y1
T3 = nC2xn-2y2
------------------------
------------------------
------------------------
Tr = nCr-1 xn-(r-1) yr-1
Putting r = r+1 in this expression, we get:
General Term: Tr+1 = nCr xn-r yr
Note : ‘Tr’ can be used as general terms also.
Problem based on General Terms
Type : 1 Find the 7th term in the expansion of
[4x – (1 / 2 x)]13
13
Sol : T7 = T6+1 = C6(4x)13-6 - (1/2 x) 6
13
= C6.47x7. 1 /(26.x3)
19. 13
= C6. 28.x4
= 13!/ (6!x7!) . 28. x4
= T7 = 439296x4
Type II Find the coefficient of x-7 in the expansion of ( ax- )11
Sol GeneralTerm,Tr+1 = 11Cr(ax)11-r - (1/bx2) r
Tr+1 =(-1)r11Cr.(a11-r/br) x11-3r ------------- (i)
Putting 11 – 3r = -7
Or 3r = 18
r=6
From (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-6
Type III : Find the term independent of ‘x’
in [(3 x2 / 2) – (1/ 3x) ] 9
Sol.: 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 = 6
So, from (i), 7th term is independent of ‘x’, and its value is:
T7 = (-1)6 . 9C6. (3-3 / 23) xo
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 from
beginning.
Ex.: Find the 4th term from the end in the expansion of
[ - ]7
Sol.: 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) .2x
T5 = 70x
Hence ‘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 is
odd. So there is only one middle term i.e. [(n/2) + 1] th them is the
middle term.
So we find (Tn+1/2). th term in this case, if ‘n’ is even.
21. Case II : When ‘n’ is odd, then total number of terms in (x+y)n is
even. So there are two middle terms i.e. (n+1) /2 th and (n+3) /2 th
are 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 – ]9
Sol.: Here total no. of terms are 10 (even). So there are true middle-
terms
i.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) x19
Greatest – term in (1+x)n : If ‘Tr’ and ‘Tr+1’ be the ‘r’ th and (r+1)th
terms in the
Expansion of (1+x)n, then :
Tr+1 = nCr(1)n-r xr = nCr xr
22. And Tr = nCr-1. xr-1
So: Tr+1 / Tr = (nCr xr / nCr-1 xr-1) = (n-r+1)/r |x|
If ‘Tr+1 be the greatest term, then Tr+1 > Tr
Or 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] 9
In 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 9/4
Tr+1 / Tr = (90- 9r) / 4r
Putting Tr+1 / Tr ≥ 1
(90-9r) / 4r ≥ 1
or 90 ≥ 13 r
or r 90 / 13
or r ≤ 6 + 12 / 13
23. T6+1 or ‘T7’ is the greatest term.
‘T7’ in [1 + (3x / 2)] 9
9
T7 = 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) / 2
II- Method : (2+3x)9 = 29 [(1 + 3x) / 2] 9
= 29 [1 + 9 / 4] 9
since x = 3 / 2
Here m = | x (n + 1) / (x + 1)| = | 9/4 (9+1) / 9/4 + 1|
= 90 / 13
So greatest term in the expansion is T[m]+1 = T3+1 = T7
Now the method is same as in method (1)
Greatest Coefficient : In any binomial expansion middle-term has the
greatest.
Coefficient. So
(i) If ‘n’ is even, then greatest – coefficient = nCn/2
24. (ii) If ‘n’ is odd, then greatest – coefficients are nC(n+1)/ 2 and
n
C (n-1)/2
Properties 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)2163
Some 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 to
the 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)nCn
and from (ii): 2n = Co + C1 + C2 + --------- + Cn
Adding these egn:
2n = 2 ( Co + C2 + C4 + ---------------)
or Co + C2 + C4 + ------- = 2n-1 ------------> (ii)
Subtracting these egn:
25. 2n = 2 (C1 + C3 + C5 + --------------)
or C1 + C3 + C5 + ------- = 2n-1 ------------> (iv)
From (iii) and (iv) :
C0 + C2 + C4 + ------- = C1 + C3 + C5 + ------- = 2n-1
Ex.: Evaluate the sum of the 8C1 + 8C3 + 8C5 + 8C7
Sol.: since nC1 + nC3 + nC5 + nC7 + -------- = 2n-1
Here 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. 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
….xmnm
Where n1, n2, n3 --------, nm are non-negative integers, satisfying the
condition
n1 + n2 + -----------+nm = n
Note: 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+3-1
Sol.: Number of terms = C3-1 = 102C2
= 102 ! / (2! X 100!)
= (102 x 101) / (2 x 1) = 5151
**General term of a multinomial – theorem :
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. zn3
Putting 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
= 13063600
Greatest coefficient in the expansion of (x1 + x2 + -------- + xm)n is
= n! / (q!) m-r ( q+1!) r
Where ‘q’ is the quotient and ‘r’ is the remainder, when ‘n’ is divided
by ‘m’.
**Example: Find the greatest coefficient in the expansion of
(a + b + c + d) 15.
Sol.: Here n = 15, m = 4
15/4 is quotient 3 and remainder 3.
since q = 3 and r = 3
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+3n3
Where 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 = 7
or n2 = 4, n3 = 1 since from (i): n1 = 5
So 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. Some tips on the solution of binomial – coefficients:
(1) If the difference of the lower suffixes of binomial coefficients in
each 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 suitable
power of ‘x’ on both sides.
Case –II : If terms are alternately positive and negative
Then:
(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 suitable
power of ‘x’ on both sides.
Note : [ (Odd – number) / 2] = 8
(2) If the sum of the lower suffixes of binomial – coefficients in
each term is same.
For Ex.: C0 Cn + C1 Cn-1 + C2. Cn-2 + ------- + Cn C0
Then:
Case – 1 : If each term is positive, then
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 suitable
power 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 suitable
power 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 positive
integer.
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. **(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] 8
Sol.: (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] 8
Sol.: 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. = 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)]8
Sol.: (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 the
term independent of ‘x’ in [x – (2/x)] 10 is 1 : 32.
33. Sol.: In (1-x)2 : Tr+1 = 10
Cr (-1)r (x2)r
Putting r = 5
T6 = -10C5 x10
Coefficient of x10 = -10C5
10
In [x – (2/x)] : Tr+1 = Cr (-1)r (x)10-r (2/x)r
= (-1)r 10Cr. 2r. x10-2r
Putting 10 – 2r = 0
r = 5
So 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. Find
the 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. 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 coefficients
of ‘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. = 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) = -12
From (i), putting the value of (m-n) :
- 9 + (m + n) = 12
or m + n = 21 -----------> (ii)
eqn (i) + eqn(ii) = 2m = 24
m = 12
Q8. If the coefficients of (2r + 1)th term and (r + 2)th term in the
expansion 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 = 43Cr
According to the questions:
43 43
C2r = Cr+1
36. 2r + r + 1 = 43
or 3r = 42
r = 14
**Q9. If the coefficient of ‘4’th and ‘13’th terms in the expansion
of [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 = nC12
According to the question:
n
C3 = nC12
n = 12 + 3
n = 15
Expansion = [x2 + (1/x)]15
15
Now Tr+1 = Cr. (x2)15-r. 1/ xr
15
Tr+1 = Cr. x30-3r -------------> (i)
37. Putting :
30 – 3r = 0
r = 10
From (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 and
6th 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. Q11. Find the coefficient of xr in the expansion of [x + (1/x)] n, if
it 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) / 2
From: (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 the
expansion of
[(y + 1)/( y 2/3 – y1/3 + 1) - (y – 1) / (y – y1/2)]10 is
210.
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. (y + 1) / (y2/3 – y1/3 + 1) = y1/3 + 1
and 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 )10
In ( y1/3 – y-1/2)10,
Tr+1 = 10Cr (y1/3)10-r. (-y-1/2)r
= (-1)r 10Cr. (10-r) / 3 - r/2
Tr+1 = (-1)r 10Cr. y(20-5r) / 6
Putting (20 – 5r) / 6 = 0
or r=4
Putting 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. **Q13: x4-r occurs in the expansion of [x + (1/ x2)] 4n, prove that its
coefficients 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
4n
From (i) Tp+1 = C4(n-r) / 4. x4r
Coefficient 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)40
General Term = Tr+1 = (1+x) [40Cr (x3)r]
= 40Cr (1+x) x3r
= 40Cr (x3r + x3r + 1)
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. So
coefficient 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!)] 2n
Sol.: General Term Tr+1 = 2nCr (x)2n-r (1/x) r
= 2nCr. x2n-2r ---------> (i)
Here 2n – 2r = 0
or n = r
From (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. Q16. The 3rd, 4th and 5th terms in the expansion of (x+a)n are
respectively ‘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. or 4n – 8 = 5n – 15
n = 7
Putting 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 / 3
or a = 2x
Putting this value in egn (vi):
7
C2. x5. 4x2 = 84
or 7! / (2! x 5!)x7 = 21
(7 x 63) / 2 x7 = 21
x7 = 1
x = 1
44. Putting this value in (ix) = a = 2
Q17. Let ‘n’ be a positive integer. If the coefficients of second, third
and fourth terms in (1+x)n are in arithmetic progression, then find the
value 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)] = 1/6
or (n – 1 – 1) / [(n-1) (n-2)] = 1/6
45. or (n- 2) / [(n-1) (n-2)] = 1/6
or 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 is
5600. Prove that x =10.
8
Sol.: T6 = T5+1 = C5 (1/ x8/3) 8-5 ( x2 log10x ) 5
or 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 = 5600
or x2(log10x)5 = 100 = 102
Clearly x = 10 satisfied as log1010 = 1.
If x > 10 or < 10, the result will change in inequality.
46. Q1 What is the 10th term in the expansion of (x-1)11 (in decreasing
powers of x)?
A
-x
B
-11x
C
-x2
D
-55x2
ANSWER (D)
Hence, the 10th term in the expansion of
(x-1)11 would be when k = 9
Q2 What is the value of
47. (Don't worry if you get it wrong ... you can learn from
your mistakes.)
A
36
B
-16
C
16
D
Zero
ANSWER (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, and
these would cancel with the second, fourth and sixth terms of (1+i)6.
Hence, the sum of the two expansions would be
i is the square root of -1,
48. A
8
B
7
C
-7
D
-8
Q 4 Use the binomial theorem formula to determine the fourth term in the
expansion
Answer
Application of the Binomial Theorem
For situations involving distribution of a net charge over an extended
region, the calculated electric field dependence may be checked in
the limit where the point of evaluation is far from the charge
distribution. 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. that amount. Often the binomial theorem is useful in considering
that limit:
The binomial theorem is especially useful in converting negative or
fractional exponents into ordinary polynomial expressions from which
the leading-order dependence may be determined. For example, for
small x the binomial theorem can be used on the following fractions
to determine the linear dependence on x:
50. Multiple angle identities
For the complex numbers the binomial theorem can be combined with
De Moivre's formula to yield multiple-angle formulas for the sine and
cosine. According to De Moivre's formula,
Using the binomial theorem, the expression on the right can be
expanded, and then the real and imaginary parts can be taken to yield
formulas for cos(nx) and sin(nx). For example, since
51. De Moivre's formula tells us that
which are the usual double-angle identities. Similarly, since
De Moivre's formula yields
In general,
and
Series for e
The number e is often defined by the formula
Applying the binomial theorem to this expression yields the usual
infinite series for e. In particular:
The kth term of this sum is
52. As n → ∞, the rational expression on the right approaches one, and
therefore
This indicates that e can be written as a series:
Indeed, since each term of the binomial expansion is an increasing
function 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, because
the powers of a and b are all 1, and only the coefficients remain.
Naturally, the values of the coefficients are not changed by the
values 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 even
powers of b will be positive and the odd powers will be negative.
