# Class 11 NCERT Solutions – Chapter 8 Binomial Theorem – Exercise 8.2

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

**Solution: **

The (r+1)

^{th}term of (x+3)^{8}is given by T_{r+1}=^{8}C_{r}(x)^{8-r}(3)^{r}(eq1).Therefore for x

^{5}we need to get 8-r =5 (Because we need to find x^{5}. Therefore, power ox must be equal to 5)So we get r=3.

Now, put r=3 in eq1. We get,

Coefficient of x

^{5}=^{8}C_{3}(x)^{5}(3)^{3}= 8!*3^{3}/(4!*4!) = 1512Coefficient of x

^{5}is 1512.

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

**Solution: **

The (r+1)

^{th}term of (a-2b)^{12}is given by T_{r+1}=^{12}C_{r}(a)^{12-r}(-2b)^{r}Ã— (eq1)In the question it is given that exponent of b is 7. Therefore, r should be equal to 7.

By putting r=7 in eq1, we get

Coefficient of a

^{5}b^{7 }=^{12}C_{7}(-2)^{7}= -101376

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

**Solution: **

General term of the equation (a+b)

^{n}is given as T_{r+1}=^{n}C_{r}(a)^{n-r}(b)^{r}.In this question a= x

^{2 }and b=-y. After putting the value of a, b and n, we get the general term asT

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

**Question 4.** Write the general term in the expansion of (x^{2}-yx)^{12}.

**Solution:**

General term of the equation (a+b)

^{n }is given as T_{r+1}=^{n}C_{r}(a)^{n-r}(b)^{r}.In this question a= x

^{2 }and b=-yx. After putting the value of a, b and n, we get the general term asT

_{r+1}=^{12}C_{r}(x^{2})^{(12-r)}(-yx)^{r}= (-1)^{r}^{12}C_{r}(x)^{(24-2r)}(y)^{r}(x)^{r }= (-1)^{r 12}C_{r}(x)^{(24-r)}(y)^{r }

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

**Solution:**

General term in the expansion of (a+b)

^{n }is written as T_{r+1}=^{n}C_{r}(a)^{n-r}(b)^{r}In the question we are given that a = x, b = -2y and n=12.

To get the 4

^{th }term, we need to put r = 3 (Because r+1=4, therefore r=3).Therefore, T

_{4}=^{12}C_{3}(x)^{12-3}(-2y)^{3}= âˆ’1760x^{9}y^{3}

**Question 6. **Find the 13^{th} term in the expansion of (9x-1/3âˆšx)^{18}

**Solution: **

General term in the expansion of (a+b)

^{n }is written as T_{r+1}=^{n}C_{r}(a)^{n-r}(b)^{r}In this question a = 9x, b= -1/3âˆšx and n=18.

To get the 13

^{th}term, we need to put r=12 (Because r+1=13, therefore r=12).Therefore, T

_{13}=^{18}C_{12}(9x)^{18-12}(-1/3âˆšx)^{12}= 18564

**Question 7. **Find the middle terms in the expansion of (3âˆ’x^{3}/6)^{7}.

**Solution: **

In the expansion of (a+b)

^{n}, if n is odd, then there are two middle terms, namely, ((n+1)/2)^{th}and ((n+1)/2+1)^{th}term.Therefore, middle terms in the expansion of (3âˆ’x

^{3}/6)^{7}are 4^{th}term and 5^{th}term.T

_{4}= T_{3+1}=^{7}C_{3}(3)^{7-3}(âˆ’x^{3}/6)^{3}= (-1)^{3}7!3^{4}x^{9}/4!.3!6^{3}=-105x^{9}/8T

_{5}= T_{4+1 }=^{7}C_{4}(3)^{7-4}(âˆ’x^{3}/6)^{4}= (-1)^{4}7!3^{3}x^{12}/4!3!6^{4}= 35x^{12}/48Thus, the middle terms are -105x

^{9}/8 and 35x^{12}/48.

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

**Solution: **

In the expansion of (a+b)

^{n},if n is even, then the middle term is (n/2+1)^{th}term.Therefore, the middle term is 6

^{th}term.T

_{6}= T_{5+1}=^{10}C_{5}(x/3)^{5}(9y)^{5}= (10!x^{5}9^{5}y^{5})/(5!5!3^{5}) = 61236x^{5}y^{5}Thus, the middle term is 61236x

^{5}y^{5}

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

**Solution: **

Let us assume that a

^{m}occurs in the (r+1)^{th}term, we obtainT

_{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=mTherefore, the coefficient of a

^{m}is^{m+n}C_{m}= (m+n)!/m!n! …(1)Let us assume that a

^{n}occurs in the (k+1)^{th}, we obtainT

_{k+1}=^{m+n}C_{k}(1)^{m+n-k}(a)^{k}=^{m+n}C_{k}a^{k}Comparing the indices of a in a

^{n }and T_{k+1}, we obtain k-nTherefore, the coefficient of a

^{n }is^{m+n}C_{n}= (m+n)!/m!n! ….(2)Thus, from (1) and (2), it can be obtained that the coefficients of a

^{m}and a^{n}are equal.

**Question 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 of 1:3:5. Find n and r.

**Solution: **

(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 (r-1)

^{th }, r^{th}and (r+1)^{th}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 of 1:3:5, we obtain

^{n}C_{r-2}/^{n}C_{r-1 }= 1/3 and^{n}C_{r-1}/^{n}C_{r}= 3/5Solving these two equations we get n-4r+5=0 and 3n-8r+3=0.

After solving these two equations, we get n=7 and r=3

Thus n=7 and r=3.

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

**Solution: **

In the expansion of (1+x)

^{2n}, T_{n+1}=^{2n}C_{n}(1)^{2n-n}(x)^{n}=^{2n}C_{n}x^{n}Therefore, coefficient of x

^{n}in the expansion of (1+x)^{2n}is^{2n}C_{n}

^{2n}C_{n}= (2n)!/(n!)^{2}….(1)Similarly, coefficient of x

^{n}in the expansion of (1+x)^{2n-1}is^{2n-1}C_{n}

^{2n-1}C_{n}= (2n)!/2.(n!)^{2}…(2)From (1) and (2), 2.

^{2n-1}C_{n}=^{2n}C_{n}Hence, it is proved that the coefficient of x

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

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

**Solution: **

General term T

_{r+1}=^{m}C_{r}(1)^{m-r}(x)^{r}=^{m}C_{r}(x)^{r}Comparing the indices of x in x

^{2}and T_{r+1}, we get r=2Therefore,

^{m}C_{2}= 6= 6

m!/(m-2)!=12

m(m-1) = 12=> m

^{2}-m-12 = 0(m-4)(m+3) = 0

m cannot be negative. Therefore, m=4

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