NCERT Solutions have been carefully compiled and developed as per the latest CBSE syllabus. The NCERT Solutions contain detailed step-by-step explanation of all the problems that are present in the textbook. The Exercise 8.2 of NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem is based on the topic General and Middle Terms. In the binomial expansion of (a + b)n , we observe that the first term is nC0an , the second term is nC1an–1b, the third term is nC2an–2b2 , and so on. Looking at the pattern of the successive terms we can say that the (r + 1)th term is nCran–rbr . The (r + 1)th term is also called the general term of the expansion (a + b)n . It is denoted by Tr+1. Thus Tr+1 = nCran–rbr.
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Download PDF of NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem Exercise 8.2
Access NCERT Solutions for Class 11 Maths Chapter 8 – Exercise 8.2
Find the coefficient of
1. x5 in (x + 3)8
Solution:
The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br
Here x5 is the Tr+1 term so a= x, b = 3 and n =8
Tr+1 = 8Cr x8-r 3r…………… (i)
For finding out x5
We have to equate x5= x8-r
⇒ r= 3
Putting value of r in (i) we get
= 1512 x5
Hence the coefficient of x5= 1512
2. a5b7 in (a – 2b)12 .
Solution:
The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br
Here a = a, b = -2b & n =12
Substituting the values, we get
Tr+1 = 12Cr a12-r (-2b)r………. (i)
To find a5
We equate a12-r =a5
r = 7
Putting r = 7 in (i)
T8 = 12C7 a5 (-2b)7
= -101376 a5 b7
Hence the coefficient of a5b7= -101376
Write the general term in the expansion of
3. (x2 – y)6
Solution:
The general term Tr+1 in the binomial expansion is given by
Tr+1 = n C r an-r br…….. (i)
Here a = x2 , n = 6 and b = -y
Putting values in (i)
Tr+1 = 6Cr x 2(6-r) (-1)r yr
= -1r 6cr .x12 – 2r. yr
4. (x2 – y x)12, x ≠ 0.
Solution:
The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br
Here n = 12, a= x2 and b = -y x
Substituting the values we get
Tn+1 =12Cr × x2(12-r) (-1)r yr xr
= -1r 12cr .x24 –2r. yr
5. Find the 4th term in the expansion of (x – 2y)12.
Solution:
The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br
Here a= x, n =12, r= 3 and b = -2y
By substituting the values we get
T4 = 12C3 x9 (-2y)3
= -1760 x9 y3
6. Find the 13th term in the expansion of
Solution:
Find the middle terms in the expansions of
Solution:
Solution:
9. In the expansion of (1 + a)m+n, prove that coefficients of am and an are equal.
Solution:
We know that the general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br
Here n= m+n, a = 1 and b= a
Substituting the values in the general form
Tr+1 = m+n Cr 1m+n-r ar
= m+n Cr ar…………. (i)
Now we have that the general term for the expression is,
Tr+1 = m+n Cr ar
Now, For coefficient of am
Tm+1 = m+n Cm am
Hence, for coefficient of am, value of r = m
So, the coefficient is m+n C m
Similarly, Coefficient of an is m+n C n
10. The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r.
Solution:
The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br
Here the binomial is (1+x)n with a = 1 , b = x and n = n
The (r+1)th term is given by
T(r+1) = nCr 1n-r xr
T(r+1) = nCr xr
The coefficient of (r+1)th term is nCr
The rth term is given by (r-1)th term
T(r+1-1) = nCr-1 xr-1
Tr = nCr-1 xr-1
∴ the coefficient of rth term is nCr-1
For (r-1)th term we will take (r-2)th term
Tr-2+1 = nCr-2 xr-2
Tr-1 = nCr-2 xr-2
∴ the coefficient of (r-1)th term is nCr-2
Given that the coefficient of (r-1)th, rth and r+1th term are in ratio 1:3:5
Therefore,
By cross multiplication
⇒ 5r = 3n – 3r + 3
⇒ 8r – 3n – 3 =0………….2
We have 1 and 2 as
n – 4r ± 5 =0…………1
8r – 3n – 3 =0…………….2
Multiplying equation 1 by number 2
2n -8r +10 =0……………….3
Adding equation 2 and 3
2n -8r +10 =0
-3n – 8r – 3 =0
⇒ -n = -7
n =7 and r = 3
11. Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n – 1.
Solution:
The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br
The general term for binomial (1+x)2n is
Tr+1 = 2nCr xr …………………..1
To find the coefficient of xn
r = n
Tn+1 = 2nCn xn
The coefficient of xn = 2nCn
The general term for binomial (1+x)2n-1 is
Tr+1 = 2n-1Cr xr
To find the coefficient of xn
Putting n = r
Tr+1 = 2n-1Cr xn
The coefficient of xn = 2n-1Cn
We have to prove
Coefficient of xn in (1+x)2n = 2 coefficient of xn in (1+x)2n-1
Consider LHS = 2nCn
12. Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.
Solution:
The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br
Here a = 1, b = x and n = m
Putting the value
Tr+1 = m Cr 1m-r xr
= m Cr xr
We need coefficient of x2
∴ putting r = 2
T2+1 = mC2 x2
The coefficient of x2 = mC2
Given that coefficient of x2 = mC2 = 6
⇒ m (m – 1) = 12
⇒ m2– m – 12 =0
⇒ m2– 4m + 3m – 12 =0
⇒ m (m – 4) + 3 (m – 4) = 0
⇒ (m+3) (m – 4) = 0
⇒ m = – 3, 4
We need positive value of m so m = 4
Access Other Exercise Solutions of Class 11 Maths Chapter 8- Binomial Theorem
Exercise 8.1 Solutions 14 Questions
Miscellaneous Exercise On Chapter 8 Solutions 10 Questions
