# CBSE Class 11 Maths – Chapter 8 Binomial Theorem- Study Materials

### CBSE Class 11 Maths Notes Chapter 8 Binomial Theorem

Binomial Expression
An expression consisting of two terms, connected by + or – sign is called binomial expression.

Binomial Theorem
If a and b are real numbers and n is a positive integer, then

The general term of (r + 1)th term in the expression is given by
Tr+1 = nCr an-r br

Some Important Observations from the Binomial Theorem
The total number of terms in the binomial expansion of (a + b)n is n + 1.

The sum of the indices of a and b in each term is n.

The coefficient of terms equidistant from the beginning and the end are equal. These coefficients are known as the binomial coefficient and
nCr = nCn-r, r = 0, 1, 2, 3,…, n

The values of the binomial coefficient steadily increase to a maximum and then steadily decrease.

The coefficient of xr in the expansion of (1 + x)n is nCr.

In the binomial expansion (a + b)n, the rth term from the end is (n – r + 2)th term from the beginning.

Middle Term in the Expansion of (a + b)n
If n is even, then in the expansion of (a + b)n, the middle term is (n/2 + 1) th term.

If n is odd, then in the expansion of (a + b)n, the middle terms are ((n+1)/2)th term and ((n+3)/2)th term.

### Binomial Theorem

• An algebraic expression containing two terms is called binomial expression.
Example
$( x+a)$
$(\frac {1}{2} +x)$
$(\frac {2}{x} – \frac{1}{x^3})$
• The general  form of the binomial expression is  $(x+a)$ and expansion $(x+a)^n , n \in N$ is called Binomial expression
• It was developed by Sir Issac newton
• The general expression for the Binomial Theorem is
$(x+a)^n = ^{n}C_{0} x^n a^0 + ^{n}C_{1} x^{n-1} a^1 + ^{n}C_{2} x^{n-2} a^2 + …..+ ^{n}C_{r} x^{n-r} a^r ….+^{n}C_{n} x^{0} a^n$
$(x+a)^{n} = \sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^{k}$
Proof:
We can prove this theorem with the help of mathematical induction
Let us assume P(n) be the statement is
$(x+a)^n = ^{n}C_{0} x^n a^0 + ^{n}C_{1} x^{n-1} a^1 + ^{n}C_{2} x^{n-2} a^2 + …..+ ^{n}C_{r} x^{n-r} a^r ….+ ^{n}C_{n} x^{0} a^n$
Step 1
Now the value of P(1)
$(x+a)^1 = ^{1}C_{0} x^1 a^0 + ^{1}C_{1} x^{1-1} a^1$
$=(x+a)$
So P(1) is true
Step 2
Now the value of P(m)
$(x+a)^m = ^{m}C_{0} x^m a^0 + ^{m}C_{1} x^{m-1} a^1 + ^{m}C_{2} x^{m-2} a^2 + …+ ^{m}C_{m} x^{0} a^m$
Now we have to prove
$(x+a)^{m+1} = ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 +…+ ^{m+1}C_{m+1} x^{0} a^{m+1}$
Now
$(x+a)^{m+1} =(x+a)(x+a)^m$
$= (x+a)( ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 + …+ ^{m+1}C_{m+1} x^{0} a^{m+1}$
$= ^m C_0 x^{m+1}a{0} + ( ^mC_1 + ^mC_0) x^ma^1 + (^mC_2 + ^mC_1) x^{m-1}a^2 +….$ $+(^m C_{m-1} + ^mC_{m}) x^1a^m + ^mC_m x^0a^{m+1}$
As $^mC_{r-1} + ^mC_r = ^{m+1}C_r$
So,
$= ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 + …+ ^{m+1}C_{m+1} x^{0} a^{m+1}$
So by principle of Mathematical induction, P(n) is true for $n \in N$

### Important conclusion from Binomial Theorem

 1 $(x+a)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^{k}$We can easily see that $(x+a)^n$ has $(n+1)$ terms as k can have values from 0 to n 2 The sum of indices of x and a in each is equal to n$x^{n-k}a^{k}$ 3 The coefficient nCr is each term is called binomial coefficient 4 $(x-a)^n$ can be treated as $[x+(-a)]^n$So$(x -a)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{n-k}a^k$So the terms in the expansion are alternatively positive and negative. The  last term is positive or negative depending on the values of n 5 $(1+x)^n$ can be treated as $(x+a)^1$ where x=1 and a=xSo$(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{k}$This is the expansion is ascending order of  power of x 6 $(1+x)^n$ can be treated as $(x+a)^n$ where x=x and a=1So$(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}$This is the expansion is descending  order of  power of x 7 $(1-x)^n$ ;can be treated as $(x+a)^n$ where x=1 and a=-xSo$(1 -x)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{k}$This is the expansion is ascending order of  power of x
Question -1
Expand using Binomial Theorem
$(x^2 +2)^5$
Solution
We know from binomial Theorem
$(x+a)^n = ^{n}C_{0} x^n a^0 + ^{n}C_{1} x^{n-1} a^1 + ^{n}C_{2} x^{n-2} a^2 + …..+ ^{n}C_{r} x^{n-r} a^r ….+^{n}C_{n} x^{0} a^n$
So putting values $x=x^2 ,a=2 \; and \; n=5$
We get
$(x^2 +2)^5= ^{5}C_{0} (x^2)^5 + ^{5}C_{1} (x^2)^4 2^1 + ^{5}C_{2} (x^2)^3 2^2 + ^{5}C_{3} (x^2)^2 2^3 + ^{5}C_{4} (x^2)^1 2^4 + ^{5}C_{5} (x^2)^0 2^5$ $=x^10 +20x^8 +160x^6 +640x^4 +1280x^2 +1024$

Practice Questions
• $(x +2)^6$
• $(1 -x^2)^5$
• $(1 -x)^7$
• $(z – x)^5$

### General Term in Binomial Expansion

$(x+a)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^k$
The First term would be = $^{n} C_0 x^n a^0$
The Second term would be =$^{n} C_1 x^{n-1} a^1$
The Third term would be = $^{n} C_2 x^{n-2} a^2$
The Fourth term would be = $^{n} C_3 x^{n-3} a^3$
Like wise (k+1) term would be
$T_{k+1}= ^n C_k x^{n-k}a^k$
This is called the general term also as every term can be find using this term
$T_1= T_{0+1}=^n C_0 x^n a^0$
$T_2= T_{1+1}=^n C_1 x^{n-1} a^1$
Similarly for
$(x -a)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{n-k}a^k$
$T_{k+1}= ^n C_k (-1)^k x^{n-k}a^k$
Again similarly for
$(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{k}$
$T_{k+1}= ^n C_k x^{k}$
Again similarly for
$(1 -x)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{k}$
$T_{k+1}= ^n C_k (-1)^k x^{k}$
To summarize it

 Binomial term (k+1) term $(x+a)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^k$ $T_{k+1}= ^n C_k x^{n-k}a^k$ $(x -a)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{n-k}a^k$ $T_{k+1}= ^n C_k (-1)^k x^{n-k}a^k$ $(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{k}$ $T_{k+1}= ^n C_k x^{k}$ $(1 -x)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{k}$ $T_{k+1}= ^n C_k (-1)^k x^{k}$

### Middle Term in Binomial Expansion

• A binomial expansion contains $(n+1)$ terms
• If n is even then the middle term would $[(\frac {n}{2}+1 ]$ th term
• If n is odd,then $\frac{n+1}{2}$ and $\frac {n+3}{3}$ are the middle term

### Solved Examples

Question-1
If the coefficient of $(2k + 4)$ and $(k – 2)$ terms in the expansion of $(1+x)^{24}$ are equal then find the value of k
Solution:
The general term of $(1 + x)^n$ is
$T_{k+1}= ^n C_k x^{k}$
Hence coefficient of (2k + 4)th term will be
$T_{2k+4} = T_{2k+3+1} = ^{24} C_{2k+3}$
and coefficient or (k – 2)th term will be
$T_{k-2} = T_{k-3+1} = ^{24} C_{k-3}$
As per question both the terms are equal
24C2k+3 = 24Ck-3.
or (2k + 3) + (k-3) = 24
k = 8
Practice Question
• Prove that $^nC_0 + ^nC_1 + ^nC_2 + …..+ ^nC_n = 2^n$
• Find the Coefficent of $x^5$ in the expansion $(1+x)^3 (1-x)^6$
• Use Binomial theorem to evaluate $(96)^3$

### Binomial Theorem Class 11 MCQs Questions with Answers

Question 1.
The coefficient of y in the expansion of (y² + c/y)5 is
(a) 10c
(b) 10c²
(c) 10c³
(d) None of these

Hint:
Given, binomial expression is (y² + c/y)5
Now, Tr+1 = 5Cr × (y²)5-r × (c/y)r
= 5Cr × y10-3r × Cr
Now, 10 – 3r = 1
⇒ 3r = 9
⇒ r = 3
So, the coefficient of y = 5C3 × c³ = 10c³

Question 2:
(1.1)10000 is _____ 1000
(a) greater than
(b) less than
(c) equal to
(d) None of these

Hint:
Given, (1.1)10000 = (1 + 0.1)10000
10000C0 + 10000C1 × (0.1) + 10000C2 ×(0.1)² + other +ve terms
= 1 + 10000×(0.1) + other +ve terms
= 1 + 1000 + other +ve terms
> 1000
So, (1.1)10000 is greater than 1000

Question 3.
The fourth term in the expansion (x – 2y)12 is
(a) -1670 x9 × y³
(b) -7160 x9 × y³
(c) -1760 x9 × y³
(d) -1607 x9 × y³

Answer: (c) -1760 x9 × y³
Hint:
4th term in (x – 2y)12 = T4
= T3+1
= 12C3 (x)12-3 ×(-2y)³
= 12C3 x9 ×(-8y³)
= {(12×11×10)/(3×2×1)} × x9 ×(-8y³)
= -(2×11×10×8) × x9 × y³
= -1760 x9 × y³

Question 4.
If n is a positive integer, then (√3+1)2n+1 + (√3−1)2n+1 is
(a) an even positive integer
(b) a rational number
(c) an odd positive integer
(d) an irrational number

Hint:
Since n is a positive integer, assume n = 1
(√3+1)³ + (√3−1)³
= {3√3 + 1 + 3√3(√3 + 1)} + {3√3 – 1 – 3√3(√3 – 1)}
= 3√3 + 1 + 9 + 3√3 + 3√3 – 1 – 9 + 3√3
= 12√3, which is an irrational number.

Question 5.
If the third term in the binomial expansion of (1 + x)m is (-1/8)x² then the rational value of m is
(a) 2
(b) 1/2
(c) 3
(d) 4

Hint:
(1 + x)m = 1 + mx + {m(m – 1)/2}x² + ……..
Now, {m(m – 1)/2}x² = (-1/8)x²
⇒ m(m – 1)/2 = -1/8
⇒ 4m² – 4m = -1
⇒ 4m² – 4m + 1 = 0
⇒ (2m – 1)² = 0
⇒ 2m – 1 = 0
⇒ m = 1/2

Question 6.
The greatest coefficient in the expansion of (1 + x)10 is
(a) 10!/(5!)
(b) 10!/(5!)²
(c) 10!/(5! × 4!)²
(d) 10!/(5! × 4!)

Hint:
The coefficient of xr in the expansion of (1 + x)10 is 10Cr and 10Cr is maximum for
r = 10/ = 5
Hence, the greatest coefficient = 10C5
= 10!/(5!)²

Question 7.
The coefficient of xn in the expansion of (1 – 2x + 3x² – 4x³ + ……..)-n is
(a) (2n)!/n!
(b) (2n)!/(n!)²
(c) (2n)!/{2×(n!)²}
(d) None of these

Hint:
We have,
(1 – 2x + 3x² – 4x³ + ……..)-n = {(1 + x)-2}-n
= (1 + x)2n
So, the coefficient of xnC3 = 2nCn = (2n)!/(n!)²

Question 8.
The value of n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively is
(a) 2
(b) 4
(c) 6
(d) 8

Hint:
Given that the first three terms of the expansion are 729, 7290 and 30375 respectively.
Now T1 = nC0 × an-0 × b0 = 729
⇒ an = 729 ……………. 1
T2 = nC1 × an-1 × b1 = 7290
⇒ n
an-1 × b = 7290 ……. 2
T3 = nC2 × an-2 × b² = 30375
⇒ {n(n-1)/2}
an-2 × b² = 30375 ……. 3
Now equation 2/equation 1
n
an-1 × b/an = 7290/729
⇒ n×b/n = 10 ……. 4
Now equation 3/equation 2
{n(n-1)/2}
an-2 × b² /n
an-1 × b = 30375/7290
⇒ b(n-1)/2a = 30375/7290
⇒ b(n-1)/a = (30375×2)/7290
⇒ bn/a – b/a = 60750/7290
⇒ 10 – b/a = 6075/729 (60750 and 7290 is divided by 10)
⇒ 10 – b/a = 25/3 (6075 and 729 is divided by 243)
⇒ 10 – 25/3 = b/a
⇒ (30-25)/3 = b/a
⇒ 5/3 = b/a
⇒ b/a = 5/3 …………….. 5
Put this value in equation 4, we get
n × 5/3 = 10
⇒ 5n = 30
⇒ n = 30/5
⇒ n = 6
So, the value of n is 6

Question 9.
If α and β are the roots of the equation x² – x + 1 = 0 then the value of α2009 + β2009 is
(a) 0
(b) 1
(c) -1
(d) 10

Hint:
Given, x² – x + 1 = 0
Now, by Shridharacharya formula, we get
x = {1 ± √(1 – 4×1×1) }/2
⇒ x = {1 ± √(1 – 4) }/2
⇒ x = {1 ± √(-3)}/2
⇒ x = {1 ± √(3 × -1)}/2
⇒ x = {1 ± √3 × √-1}/2
⇒ x = {1 ± i√3}/2 {since i = √-1}
⇒ x = {1 + i√3}/2, {–1 – i√3}/2
⇒ x = -{-1 – i√3}/2, -{-1 + i√3}/2
⇒ x = w, w² {since w = {-1 + i√3}/2 and w² = {-1 – i√3}/2 }
Hence, α = -w, β = w²
Again we know that w³ = 1 and 1 + w + w² = 0
Now, α2009 + β2009 = α2007 × α² + β2007 × β²
= (-w)2007 × (-w)² + (-w²)2007 × (-w²)² {since 2007 is multiple of 3}
= -(w)2007 × (w)² – (w²)2007 × (w4)
= -1 × w² – 1 × w³ × w
= -1 × w² – 1 × 1 × w
= -w² – w
= 1 {since 1 + w + w² = 0}
So, α2009 + β2009 = 1

Question 10.
The general term of the expansion (a + b)n is
(a) Tr+1 = nCr × ar × br
(b) Tr+1 = nCr × ar × bn-r
(c) Tr+1 = nCr × an-r × bn-r
(d) Tr+1 = nCr × an-r × br

Answer: (d) Tr+1 = nCr × an-r × br
Hint:
The general term of the expansion (a + b)n is
Tr+1 = nCr × an-r × br

Question 11.
The coefficient of xn in the expansion (1 + x + x² + …..)-n is
(a) 1
(b) (-1)n
(c) n
(d) n+1

Hint:
We know that
(1 + x + x² + …..)-n = (1 – x)-n
Now, the coefficient of x = (-1)n × nCn
= (-1)n

Question 12.
If n is a positive integer, then (√5+1)2n + 1 − (√5−1)2n + 1 is
(a) an odd positive integer
(b) not an integer
(c) none of these
(d) an even positive integer

Hint:
Since n is a positive integer, assume n = 1
(√5+1)² + 1 − (√5−1)² + 1
= (5 + 2√5 + 1) + 1 – (5 – 2√5 + 1) + 1 {since (x+y)² = x² + 2xy + y²}
= 4√5 + 2, which is not an integer

Question 13.
In the expansion of (a + b)n, if n is even then the middle term is
(a) (n/2 + 1)th term
(b) (n/2)th term
(c) nth term
(d) (n/2 – 1)th term

Answer: (a) (n/2 + 1)th term
Hint:
In the expansion of (a + b)n,
if n is even then the middle term is (n/2 + 1)th term

Question 14.
In the expansion of (a + b)n, if n is odd then the number of middle term is/are
(a) 0
(b) 1
(c) 2
(d) More than 2

Hint:
In the expansion of (a + b)n,
if n is odd then there are two middle terms which are
{(n + 1)/2}th term and {(n+1)/2 + 1}th term

Question 15.
if n is a positive ineger then 23nn – 7n – 1 is divisible by
(a) 7
(b) 9
(c) 49
(d) 81

Hint:
Given, 23n – 7n – 1 = 23×n – 7n – 1
= 8n – 7n – 1
= (1 + 7)n – 7n – 1
= {nC0 + nC1 7 + nC2 7² + …….. + nCn 7n} – 7n – 1
= {1 + 7n + nC2 7² + …….. + nCn 7n} – 7n – 1
= nC2 7² + …….. + nCn 7n
= 49(nC2 + …….. + nCn 7n-2)
which is divisible by 49
So, 23n – 7n – 1 is divisible by 49

Question 16.
In the binomial expansion of (71/2 + 51/3)37, the number of integers are
(a) 2
(b) 4
(c) 6
(d) 8

Hint:
Given, (71/2 + 51/3)37
Now, general term of this binomial Tr+1 = 37Cr × (71/2)37-r × (51/3)r
⇒ Tr+1 = 37Cr × 7(37-r)/2 × (5)r/3
This General term will be an integer if 37Cr is an integer, 7(37-r)/2 is an integer and (5)r/3 is an integer.
Now, 37Cr will always be a positive integer.
Since 37Cr denotes the number of ways of selecting r things out of 37 things, it can not be a fraction.
So, 37Cr is an integer.
Again, 7(37-r)/2Cr will be an integer if (37 – r)/2 is an integer.
So, r = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37 …………. 1
And if (5)r/3 is an integer, then r/3 should be an integer.
So, r = 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36 ………….2
Now, take intersection of 1 and 2, we get
r = 3, 9, 15, 21, 27, 33
So, total possible value of r is 6
Hence, there are 6 integers are in the binomial expansion of (71/2 + 51/3)37

Question 17.
The number of ordered triplets of positive integers which are solution of the equation x + y + z = 100 is
(a) 4815
(b) 4851
(c) 8451
(d) 8415

Hint:
Given, x + y + z = 100;
where x ≥ 1, y ≥ 1, z ≥ 1
Let u = x – 1, v = y – 1, w = z – 1
where u ≥ 0, v ≥ 0, w ≥ 0
Now, equation becomes
u + v + w = 97
So, the total number of solution = 97+3-1C3-1
= 99C2
= (99 × 98)/2
= 4851

Question 18.
The greatest coefficient in the expansion of (1 + x)10 is
(a) 10!/(5!)
(b) 10!/(5!)²
(c) 10!/(5! × 4!)²
(d) 10!/(5! × 4!)

Hint:
The coefficient of xr in the expansion of (1 + x)10 is 10Cr and 10Cr is maximum for
r = 10/2 = 5
Hence, the greatest coefficient = 10C5
= 10!/(5!)²

Question 19.
If the third term in the binomial expansion of (1 + x)m is (-1/8)x² then the rational value of m is
(a) 2
(b) 1/2
(c) 3
(d) 4

Hint:
(1 + x)m = 1 + mx + {m(m – 1)/2}x² + ……..
Now, {m(m – 1)/2}x² = (-1/8)x²
⇒ m(m – 1)/2 = -1/8
⇒ 4m² – 4m = -1
⇒ 4m² – 4m + 1 = 0
⇒ (2m – 1)² = 0
⇒ 2m – 1 = 0
⇒ m = 1/2

Question 20.
In the binomial expansion of (a + b)n, the coefficient of fourth and thirteenth terms are equal to each other, then the value of n is
(a) 10
(b) 15
(c) 20
(d) 25