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

Chapter 8 Binomial Theorem Exercise 8.2 Class 11 Maths NCERT Solutions is provided here that will essential for development of problem solving skills. NCERT Solutions for Class 11 Maths will help every students in completing their home assignments on time and easily. These NCERT Solutions are updated according to the latest pattern released by CBSE which will save your precious time.

1. Find the coefficient of x 5 in (x + 3)8.Â

**Answer**

^{8}Â =Â

^{8}C

_{r}Â x

^{8 â€“ r}. 3

^{r}

We have to find the coefficient of x

^{Â 5}

8 â€“ r = 5, r = 8 â€“ 5 = 3

âˆ´Â Coefficient of x

^{Â 5}Â (putting r = 3)

=Â

^{8}C

_{3}. 3

^{3}Â = (8.7.6)/( 1.2.3) . 27 = 56 . 27 = 1512.

2. Find the coefficient of a

^{5}bÂ

^{7}Â in (a â€“ 2b)

^{12}.

**Answer**

(a â€“ 2b)

^{Â 12}Â = [a + (â€“ 2b)]

^{12}

General term T

_{r + 1Â }= C(12, r) a

^{Â 12 â€“ r}Â (â€“2b)

^{Â r}Â .

Putting 12 â€“ r = 5 or 12 â€“ 5 = r => r = 7

T

_{7 +1Â }= C (12, 7) aÂ

^{12 â€“ 7}Â (â€“2b)Â

^{7}

= C(12, 7) a

^{Â 5}Â (â€“2b)

^{Â 7}= C(12, 7) (â€“2)

^{7}Â aÂ

^{5}b

^{Â 7}

Hence required coefficient is C(12, 7) (â€“2)

^{7}

= 12!/ (7! 5!) . 2

^{7}

= (-12Â Ã—Â 11 Ã— 10 Ã— 9 Ã— 8 Ã— 7!)/(7!Â Ã—Â 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1) Ã— 2

^{7}

= 8 Ã—Â -11 Ã— 9 Ã— 2

^{7}

= -99 Ã—Â 8 Ã— 128 = -101376.

3. Write the general term in the expansion of (x

^{2}Â â€“ y)

^{6}

**Answer**

General term = T

_{r + 1Â }=Â

^{6}C

_{r}Â (xÂ

^{2})

^{6}Â

^{â€“ r}Â (â€“ y)

^{Â r}

= (â€“1)

^{r}Â 6!/(r!(6 â€“ r)!) .x

^{12-2rÂ }.y

^{rÂ }

4. Write the general term in the expansion of (x

^{2}Â â€“ yx)

^{12}, xÂ â‰ Â 0.

**Answer**

Binomial expansion is (x

^{Â 2}Â â€“ yx)

^{Â 12}

General term T

_{r +1Â }=Â

^{12}C

_{r}Â (x

^{Â 2})Â

^{12 â€“ r}. (â€“yx)Â

^{r}

= 12!/(r!(12 â€“ r)!) . x

^{24 â€“ 2rÂ }.(-1)

^{r}Â . y

^{r}Â x

^{r}

= ((-1)

^{r}Â 12!)/(r!(12 â€“ r)!) . x

^{24-rÂ }. y

^{r}

5. Find the 4

^{th}Â term in the expansion of (x â€“ 2y)

^{12}

**Answer**

4

^{th}Â term = T

_{3+1Â }in the expansion of (x+(-2y))

^{12}

=Â

^{12}C

_{3}.x

^{12 â€“ 3Â }[-2y]

^{3}

= (12 . 11. 10)/(1. 2. 3) x

^{9}Â (-1)

^{3}.2

^{3}. y

^{3}

= -220 Ã— 8x

^{9}.y

^{3}Â = -1760 x

^{9}Â y

^{3}

6. Find the 13th term in the expansion of

(9x â€“ 1/3âˆšx)

^{18}Â , xÂ â‰ Â 0

**Answer**

13

^{th}Â term, T

_{13}Â = T

_{12 +1}

=Â

^{18}C

_{12}Â (9x)

^{18 â€“ 12}Â (-(1/3âˆšx))

^{12}

=Â

^{18}C

_{6}9

^{6}x

^{6}Â (-1)

^{12}Â . 1/3

^{12}Â Ã—Â 1/x

^{6}

= 18564Â Ã— (3

^{2})

^{6}. 1/3

^{12}Ã—Â x

^{6}/x

^{6}

= 18564Â Ã— 3

^{12}/3

^{12}Â = 18564

7. Find the middle term in the expansion of (3 = x

^{3}/6)

^{7}.

**Answer**

Number of terms in the expansion is 7 + 1 = 8

There are two middle terms which are

(8/2)

^{th}Â & ((7 + 3)/2)

^{th}Â i.e., 4

^{th}Â & 5

^{th}

Hence, we have to find T

_{4}Â and T

_{5}Â in the given expansion

(3 â€“ (x

^{3}/6))

^{7}Â = [3 + (-(x

^{3}/6)]

^{7}

T

_{r+1Â }= C(7, r)3

^{7-r}Â (-(x

^{3}/6)

^{r}) â€¦â€¦â€¦(i)

Now T

_{r}Â

_{+ 1}Â = T

_{4}Â or r + 1 = 4 âˆ´Â r = 3

Putting r = 3, we have

T

_{3+1Â }= C(7, 3)3

^{7 â€“ 3}(-(x

^{3}/6)

^{3})

= C(7, 3)3

^{4}Â (-1)

^{3}Â x

^{9}/6

^{3}Â = -7!/(3!4!) . 3/2

^{3}Â . x

^{9}

= (- 7Â Ã—6Â Ã—5Â Ã—4!)/(3 Ã—Â 2 Ã— 1 Ã— 4!) . 3/2

^{3}Â . x

^{9}Â = -105/8 x

^{9}

Again T

_{r+1}Â = T

_{5}Â or r + 1= 5 or r = 4

Putting r = 4 in (i), we have

T

_{4+ 1}Â = T

_{5}Â = C(7, 4) 3Â

^{7 â€“ 4}Â (-1)

^{4 }x

^{12}/6

^{4}

= 7/4!3!Â . 3

^{3}x

^{12}/3

^{4}2

^{4}Â = (7Ã—Â 6 Ã— 5 Ã— 4!)/(3 Ã—Â 2 Ã— 1 Ã— 4!) Ã—Â x

^{12}/3 Ã— 2

^{4}

35/48 .x

^{12}

8. Find the middle term in the expansion of (x/3 + 9y)

^{10}

**Answer**

Number of terms in the expansion is

10 + 1 = 11 (odd)

Middle term of the expansions is (n/2 + 1)

^{th}Â term

= (5 + 1)

^{th}Â term = 6

^{th}Â term

T

_{6}Â = T

_{5 + 1Â }= C(10, 5)(x/3)

^{10 â€“ 5}(9y)

^{5}

= C(10, 5) x

^{5}/3

^{5}Â . 9

^{5}y

^{5Â }= C(10, 5) 3

^{5}x

^{5}y

^{5}

= 10!/(5! (10 â€“ 5)!) 3

^{5}x

^{5}y

^{5}Â = 10!/5!5! . 3

^{5}x

^{5}y

^{5}

= (10Â Ã—Â 9 Ã—8Â Ã—7Â Ã—Â 6 Ã— 5!)/(5 Ã—Â 4 Ã—3Â Ã—2Â Ã—1Ã—Â 5!) 3

^{5}x

^{5}y

^{5}

= 61236x

^{5}y

^{5}

9. In the expansion of (1 + a)Â

^{m + n}, prove that coefficients of a

^{Â m}Â and a

^{Â nÂ }are equal.

**Answer**

General term in the expansion of (1 + a)

^{Â m + nÂ }is

T

_{r + 1Â }=Â

^{m + n}C

_{r}Â a

^{Â r}

Putting r = m

T

_{m+1Â }=Â

^{m+n}C

_{m}a

^{m}â€¦â€¦(i)

âˆ´Â Coefficient of a

^{m}Â =Â

^{m + n}C

_{m}

Again putting r = n

T

_{n+1}Â =Â

^{m+n}C

_{n}a

^{n}

Coefficient of a

^{n}Â =Â

^{m+n}C

_{n}Â =Â

^{m+n}C

_{mÂ }..â€¦ (ii)

[âˆµ

^{n}C

_{r}Â =Â

^{n}C

_{n â€“ r}]

From (i) and (ii) coefficient of a

^{m}Â is equal to coefficient of a

^{n}Â .

10. The coefficients of the (r â€“ 1)

^{th}, r

^{Â th}Â and (r + 1)

^{th}Â terms in the expansion of (x + 1)

^{n}Â are in the ratio of 1: 3: 5. Find n and r.

**Answer**

General term in the expansion of (x + 1)

^{n}Â is

T

_{r â€“ 1}Â = T

_{(r â€“ 2)}Â =Â

^{n}C

_{r â€“ 2}. x

^{r-2}

T

_{r}Â = T

_{(r â€“ 1)}Â =Â

^{n}C

_{r-1}Â . xÂ

^{r -1}

T

_{r+1}Â =Â

^{n}C

_{r}.x

^{r}

C(n, r -2) : C(n, r -1): C(n, r) = 1 : 3 : 5

Or (C(n, r â€“ 2))/1 = (C(n, r â€“ 1))/3 = (C(n,r))/5

If (C(n, r -2))/1 = (C(n, r â€“ 1))/3

Or 3C(n, r- 2) = C(n, r â€“ 1)

Or 3. n!/((r â€“ 2) ! (n + 2 â€“ r) !) = n!/((r â€“ 1) ! (n + 1 â€“ r) !)

Or 3/((r â€“ 2)!(n + 2 â€“ r)(n + 1 â€“ r) !)

= 1/((r â€“ 1)(r â€“ 2)!(n + 1 â€“ r)!)

Or 3/(n + 2 â€“ r) = 1/(r â€“ 1)

Or 3r â€“ 3 = n + 2 â€“ r

Or 4r = n + 5 â€¦.(i)

Again if (C(n, r â€“ 1))/3 = (C(n,r))/5

Or 5C(n, r â€“ 1) = 3C(n, r)

Or 5. n!/((r â€“ 1) !(n + 1 â€“ r) !) = 3. n!/(r!(n â€“ r) !)

Or 5/((r â€“ 1) !(n + 1 â€“ r)(n â€“ r) ! = 3/(r(r â€“ 1)!(n â€“ r)!)

Or 5r = 3(n + 1 â€“ r) or 8r = 3n + 3 â€¦.(ii)

From (i) & (ii) 2n + 10 = 3n + 3

Or 3n â€“ 2n = 10 -3 => n = 7

From (ii) 8r = 21 + 3 = 24

âˆ´Â r = 3

âˆ´Â n = 7, r = 3

11. Prove that the coefficient of x

^{n}Â in the expansion of (1 + x)

^{2n}Â is twice the coefficient of x

^{n}Â in the expansion of (1 + x)

^{2n â€“1}Â .

**Answer**

General term in the expansion of (1 + x)

^{Â 2nÂ }is

T

_{r + 1Â }= C (2n, r) x

^{Â r}

Putting r = n, we have

T

_{n + 1Â }= C(2n, n) x

^{Â n}

Coefficient of x

^{Â n}Â = C(2n, n)

Again general term in the expansion of

(1 + x)

^{2n â€“1}Â is T

_{r + 1Â }= C (2n â€“1, r) x

^{Â r}

Putting r = n, we have

T

_{n + 1Â }= C(2n â€“ 1, n) x

^{Â n}

Coefficient of x

^{Â n}Â in the expansion of (1+ x)

^{2n â€“ 1Â }is C(2n â€“ 1, n)

According to the problem, we have to prove that

C(2n, n) = 2Â Ã—Â C(2n â€“ 1, n)

Or 2n!/(n!(2n â€“ n)!) = 2. ((2n â€“ 1)!)/(n!(2n â€“ 1 â€“ n)!)

Or 2n!/n!n! = 2. ((2n â€“ 1)!)/(n!(n â€“ 1)!)

Multiplying N

^{r}Â and D

^{r}Â by n on RHS, we have

2n!/n!n! = (2n(2n -1)!)/(n!n(n â€“ 1)!)

i.e 2n!/n!n! = 2n!/n!n!, Which is true.

Hence proved.

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

**Answer**

Given expansion is (1 + x)

^{m}. Now,

General term = T

_{r+1Â }=Â

^{m}C

_{r}x

^{r}

Put r = 2, we have

T

_{3}Â =Â

^{m}C

_{2}.x

^{2}

According to the question C(m, 2) = 6

or (m(m â€“ 1))/2! = 6

=> m

^{2}Â â€“ m = 12

Or m

^{2}Â â€“ m â€“ 12 = 0

=> m

^{2}Â â€“ 4m + 3m â€“ 12 = 0

Or (m â€“ 4)(m + 3) = 0

âˆ´Â m = 4, since mÂ â‰ â€“3