NCERT Solutions Class 11th Maths Chapter – 8 Binomial Theorem Exercise 8.2

NCERT Solutions Class 11th Maths Chapter – 8 Binomial Theorem Exercise 8.2

NCERT Solutions Class 11th Maths Chapter – 8 Binomial Theorem Exercise 8.2

?Chapter – 8?

✍Binomial Theorem✍

?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 T_r+1 = ⁿ C _r a^n-r b^r
Here x⁵ is the T_r+1 term so a= x, b = 3 and n =8
T_r+1 = ⁸C_r x^8-r 3^r…………… (i)
For finding out x⁵
We have to equate x⁵= x^8-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 byT_r+1 = ⁿ C _r a^n-r b^r
Here a = a, b = -2b & n =12
Substituting the values, we get
T_r+1 = ¹²C_r a^12-r (-2b)^r………. (i)
To find a⁵
We equate a^12-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 T_r+1 in the binomial expansion is given by
T_r+1 = ⁿ C _r a^n-r b^r…….. (i)
Here a = x² , n = 6 and b = -y
Putting values in (i)
T_r+1 = ⁶C_r x ^2(6-r) (-1)^r y^r

= -1^{r 6}c_r .x^{12 – 2r}. y^r

4. (x² – y x)¹², x ≠ 0.

?‍♂️solution – The general term Tr+1 in the binomial expansion is given by  T_r+1 = ⁿ C _r a^n-r b^r
Here n = 12, a= x² and b = -y x
Substituting the values we get
T_n+1 =¹²C_r × x^2(12-r) (-1)^r y^r x^r

= -1^{r 12}c_r .x^{24 –2r}. y^r

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 13^th 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 a^m and aⁿ are equal.

?‍♂️solution – _{We know that the general term Tr+1 in the binomial expansion is given by Tr+1 = ⁿCr a^n-r b^r
Here n= m+n, a = 1 and b= a
Substituting the values in the general form
Tr+1 = ^m+n Cr 1^m+n-r a^r
= ^m+n Cr a^r…………. (i)
Now we have that the general term for the expression is,
Tr+1 =  ^m+n Cr a^r
Now, For coefficient of a^m
Tm+1 =  ^m+n Cm a^m
Hence, for coefficient of a^m, 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, r^th 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 T_r+1 = ⁿC_r a^n-r b^r
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) = ⁿC_r 1^n-r x^r
T_(r+1) = ⁿC_r x^r
The coefficient of (r+1)^th term is ⁿC_r
The r^th term is given by (r-1)^th term
T_(r+1-1) = ⁿC_r-1 x^r-1
T_r = ⁿC_r-1 x^r-1
∴ the coefficient of r^th term is ⁿC_r-1
For (r-1)^th term we will take (r-2)^th term
T_r-2+1 = ⁿC_r-2 x^r-2
T_r-1 = ⁿC_r-2 x^r-2
∴ the coefficient of (r-1)^th term is ⁿC_r-2
Given that the coefficient of  (r-1)^th, r^th and r+1^th term are in ratio 1:3:5
Therefore,

⇒ 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 T_r+1 = ⁿC_r a^n-r b^r
The general term for binomial  (1+x)²ⁿ is
T_r+1 = ²ⁿC_r x^r …………………..1
To find the coefficient of xⁿ
r = n
T_n+1 = ²ⁿC_n xⁿ
The coefficient of  xⁿ = ²ⁿC_n
The general term for binomial (1+x)^2n-1 is
T_r+1 = ^2n-1C_r x^r
To find the coefficient of xⁿ
Putting n = r
T_r+1 = ^2n-1C_r xⁿ
The coefficient of xⁿ = ^2n-1C_n
We have to prove
Coefficient of  xⁿ in (1+x)²ⁿ = 2 coefficient of xⁿ in (1+x)^2n-1
Consider LHS =  ²ⁿC_n

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 T_r+1 = ⁿC_r a^n-r b^r
Here a = 1, b = x and n = m
Putting the value
T_r+1 = ^m C_r 1^m-r x^r
= ^m C_r x^r
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