# Class 11 Maths NCERT Solutions for Chapter 8 Binomial Theorem Exercise 8.2

### Binomial Theorem Exercise 8.2 Solutions

**1. Find the coefficient of x^{5} in (x + 3)^{8}**

**Solution**

It is known that (r + 1)^{th} term, (T_{r}_{+1}), in the binomial expansion of (a + b)^{n} is given by T_{r+1} = ^{n}C_{r} a^{n-r}b^{r}.

Assuming that x^{5} occurs in the (r + 1)^{th} term of the expansion (x + 3)^{8} , we obtain

T_{r+1} = ^{8}C_{r} x^{8-r }3^{r}.

Comparing the indices of x in x^{5} and in T_{r+1}, we obtain

r = 3

Thus, the coefficient of x^{5} is ^{8}C_{r} (3)^{3} = (8!/3!5!) × 3^{3} = (8.7.6.5!/3.2.5!).3^{3} = 1512

**2. Find the coefficient of a ^{5} b^{7} in (a - 2b)^{12} .**

**Solution**

It is known that (*r *+ 1)^{th} term, (*T*_{r}_{+1}), in the binomial expansion of (*a *+ *b*)^{n} is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r} .

Assuming that a^{5} b^{7} occurs in the (r + 1)^{th} term of the expansion (a - 2b)^{12} , we obtain

Comparing the indices of a and b in a^{5} b^{7} and in T_{r+1} , we obtain

r = 7

Thus, the coefficient of a^{5} b^{7} is ^{12}C_{7} (-2)^{7} = (-12!/7!5!) .2^{7} = (12.11.10.9.8.7!/5.4.3.2.7!) 2^{7} = -(792)(128) = -101376

**3. Write the general term in the expansion of (x ^{2} - y)^{6} **

**Solution**

It is known that the general term T_{r+1}{which is the (r+1)^{th} term} in the binomial expansion of (a + b)^{n} is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r} .

Thus, the general term in the expansion of (x^{2} - y^{6}) is

T_{r+1} = ^{6}C_{r} (x^{2} )^{6-r} (-y)^{r} . (-1)^{r} ^{6}C_{r} x^{12-2r} y^{r}

**4. Write the general term in the expansion of ( x^{2} – yx)^{12}, x ≠ 0**

**Solution**

It is known that the general term T_{r+1} {which is the (r + 1)^{th} term} in the binomial expansion of (a + b)^{2} is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r} .

Thus, the general term in the expansion of (x^{2} - yx)^{12} is

**5. Find the 4 ^{th} term in the expansion of (x – 2y)^{12 }.**

**Solution**

It is known that (r + 1)^{th} term, T_{r+1}, in the binomial expansion of (a + b)^{n} is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r} .

Thus, the 4th term in the expansion of (x - 2y)^{12} is

**6. Find the 13th term in the expansion of (9x - 1/3√x) ^{18} , x ≠ 0. **

**Solution**

It is known that (r + 1)th term, (T_{r+1} ), in the binomial expansion of (a + b)^{n} is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}.

Thus, 13th term in the expansion of (9x - 1/3√x)^{18} is

**7. Find the middle terms in the expansions of (3 - x ^{3} /6)^{7} . **

**Solution**

It is known that in the expansion of (a + b)^{n}, if n is odd, then there are two middle terms, namely, [(n+1)/2]^{8} term and [(n+1)/2 + 1)^{th} term.

Therefore, the middle terms in the expansion of [3 - x^{3} /6]^{7} are [(7+1)/2]^{th} = 4^{th} term and [(7+1)/2 + 1)^{th} = 5^{th} term

**8. Find the middle terms in the expansions of (x/3 + 9y) ^{10} **

**Solution**

It is known that in the expansion (a + b)^{n} , if n is even , then the middle term is (n/2 + 1)^{th} term.

Therefore, the middle term in the expansion of (x/3 + 9y)^{10} is (10/2 + 1)^{th} = 6th term

Thus, the middle term in the expansion of (x/3 + 9y)^{10} is 61236 x^{5} y^{5}.

**9. In the expansion of (1 + a)^{m + n}, prove that coefficients of a^{m} and a^{n} are equal.**

**Solution**

It is known that (*r *+ 1)^{th} term, (*T*_{r}_{+1}), in the binomial expansion of (*a *+ *b*)^{n} is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r} .

Assuming that *a*^{m} occurs in the (*r* + 1)^{th} term of the expansion (1 + *a*)^{m}^{ + }^{n}, we obtain

T_{r+1} = ^{m+n}C_{r} (1)^{m+n-r} (a)^{r} . = ^{m+n}C_{r} a^{r}

Comparing the indices of a in a^{m} and in T_{r+1}, we obtain

r = m

Therefore, the coefficient of a^{m} is

Assuming that a^{n} occurs in the (k+1)^{th} term of the expansion (1 + a)^{m+n}, we obtain

Comparing the indices of a in a^{n} and in T_{k+1}, we obtain

k = n

Therefore, the coefficient of a^{n} is

Thus, from (1) and (2), it can be observed that the coefficients of a^{m} and a^{n} in the expansion of (1+ a)^{m+n} are equal.

**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**

It is known that (k +1)^{th} term, (T_{k+1} ) in the binomial expansion of (a + b)^{n} is given by T_{k+1} = ^{n}C_{k} a^{n-k} b^{k} .

Therefore, (r – 1)^{th} term in the expansion of (x + 1)^{n} is T_{r-1} = ^{n}C_{r-2} (x)^{n-(r-2)} (1)^{(r- 2)} = ^{n}C_{r-2} X^{n-r+2}

r^{th} term in the expansion of (x + 1)^{n} is T_{r} = ^{n}C_{r-1} (x)^{n-(r-1)} (1)^{(r-1)} = ^{n}C_{r-1} x^{n-r+1}

(r +1)^{th} term in the expansion of (x +1)^{n} is T_{r+1} = ^{n}C_{r} (X)^{n-r} (1)^{r} = ^{n}C_{r} X^{n-r}

Therefore, the coefficients of the (r -1)^{th} , r^{th} and (r + 1)^{th} terms in the expansion of (x + 1)^{n} are ^{n}C_{r-2} , ^{n}C_{r-1} and ^{n}C_{r} respectively. Since these coefficients are in the ratio 1 : 3 : 5, we obtain

Multiplying (1) by 3 and subtracting it from (2), we obtain

4*r *– 12 = 0

⇒ *r* = 3

Putting the value of *r* in (1), we obtain*n* – 12 + 5 = 0

⇒ *n* = 7

Thus, *n *= 7 and *r* = 3

**11. Prove that the coefficient of x^{n} in the expansion of (1 + x)^{2}^{n} is twice the coefficient of x^{n} in the expansion of (1 + x)^{2}^{n}^{–1 }.**

**Solution**

It is known that (*r *+ 1)^{th} term, (*T*_{r}_{+1}), in the binomial expansion of (*a *+ *b*)^{n} is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r} ..

Assuming that *x*^{n} occurs in the (*r* + 1)^{th} term of the expansion of (1 + *x*)^{2}^{n}, we obtain

Comparing the indices of x in x^{n} and in T_{r+1} , we obtain

r = n

Therefore, the coefficient of x^{n} in the expansion of (1 + x)^{2n} is

Assuming that *x*^{n} occurs in the (*k* +1)^{th} term of the expansion (1 + *x*)^{2}^{n }^{– 1}, we obtain

Comparing the indices of x in *x*^{n} and *T*_{k}_{ + 1}, we obtain*k* =* n*Therefore, the coefficient of

*x*

^{n}in the expansion of (1 +

*x*)

^{2}

^{n }

^{–1}is

Therefore, the coefficient of x

^{n}in the expansion of (1 + x)

^{2}

^{n}is twice the coefficient of x

^{n}in the expansion of (1 + x)

^{2}

^{n}

^{–1}.

Hence, proved.

**12. Find a positive value of m for which the coefficient of x^{2} in the expansion (1 + x)^{m} is 6**

**Solution**

It is known that (*r *+ 1)^{th} term, (*T*_{r}_{+1}), in the binomial expansion of (*a *+ *b*)^{n} is given by .

T_{r+1} = ^{n}C_{r} a^{n-r} b^{r} .

Assuming that *x*^{2} occurs in the (*r *+ 1)^{th} term of the expansion (1 +*x*)^{m}, we obtain

Comparing the indices of *x* in *x*^{2} and in *T*_{r}_{ + 1}, we obtain*r* = 2

Therefore, the coefficient of *x*^{2} is ^{m}C_{2} .

It is given that the coefficient of *x*^{2} in the expansion (1 + *x*)^{m} is 6.

⇒ m(m - 1) = 12

⇒ m^{2} - m - 12 = 0

⇒ m^{2} - 4m + 3m - 12 = 0

⇒ m(m - 4) + 3(m -4) = 0

⇒ (m - 4)(m + 3) = 0

⇒ (m - 4) = 0 or (m +3) = 0

⇒ m = 4 or m = -3

Thus, the positive value of m, for which the coefficient of x^{2} in the expansion

(1 + x)^{m} is 6, is 4.