Find the 13 t h term in the expansion of 9 x - 1 3 x 18 , x ≠ 0

We know that, In the expansion of ( a + b ) n , the ( r + 1 ) t h term is given by, T r + 1 = C r n a n - r b r . Here, a = 9 x , b = 1 3 x , r = 12 and n = 18 . Therefore, the 13 t h term is given by, T 13 = T 12 + 1 = C 12 18 . ( 9 x ) 18 - 12 - 1 3 x 12 = - 1 12 . C 12 18 . 9 x 6 1 3 x 12 = 18 ! 6 ! 12 ! × 9 6 × 1 3 12 × x 6 × 1 x 6 ∵ C r n = n ! r ! ( n - r ) ! = 18 ! 6 ! 12 ! 9 6 = 3 2 6 = 3 12 = 18 × 17 × 16 × 15 × 14 × 13 × 12 ! 6 × 5 × 4 × 3 × 2 × 1 × 12 ! = 18564

Find the middle terms in the expansions of 3 - x 3 6 7 .

It is known that the general 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 . Similarly, the general term of 3 - x 3 6 7 is given by T r + 1 = 7 C r ( 3 ) 7 - r - x 3 6 r . Here, n = 7 , which is odd. So, there are two middle terms, given by 4 th term = n + 1 2 t h term and 5 t h term = n + 1 2 + 1 t h term. So, T 4 = T 3 + 1 = 7 C 3 ( 3 ) 7 - 3 - x 3 6 3 = - 7 C 3 3 4 x 9 6 3 = - 7 ! 3 ! 4 ! × 81 × x 9 216 = - 105 8 x 9 T 5 = T 4 + 1 = 7 C 4 ( 3 ) 7 - 4 - x 3 6 4 = C 4 7 3 3 x 12 6 4 = 7 ! 4 ! 3 ! × 27 × x 12 1296 = 35 48 x 12

Find the middle term in the expansions of x 3 + 9 y 10 .

To find: The middle term in the expansions of x 3 + 9 y 10 . We know that, in the expansion of ( a + b ) n , number of terms = n + 1 . So, if n is even, the middle term is the n 2 + 1 t h term. Therefore, here the middle term = 10 2 + 1 = 6 t h term. We know that the ( r + 1 ) t h in the given expansion is T r + 1 = C r n a n - r b r . So, the 6 t h term is given by, T 6 = T 5 + 1 = C 5 10 x 3 10 - 5 9 y 5 = 10 ! 5 ! 5 ! × ( x 5 3 5 ) ( 9 5 y 5 ) ∵ C r n = n ! r ! ( n - r ) ! = 10 × 9 × 8 × 7 × 6 × 5 ! 1 × 2 × 3 × 4 × 5 × 5 ! ( x 5 3 5 ) ( 3 10 y 5 ) = 252 . x 5 . 3 5 . y 5 = 61236 x 5 y 5

E M B I B E

MATHEMATICS Textbook for Class XI>Chapter 8 - Binomial Theorem>EXERCISE 8.2

EXERCISE 8.2

Author:Jharkhand Board

Jharkhand Board Mathematics Solutions for EXERCISE 8.2

Simple step-by-step solutions to EXERCISE 8.2 questions of Binomial Theorem from MATHEMATICS Textbook for Class XI. Also get 3D topic explainers, cheat sheets, and unlimited doubts solving on EMBIBE.

Questions from EXERCISE 8.2 with Hints & Solutions

EASY

1th Jharkhand Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 8 - Binomial Theorem>EXERCISE 8.2>Q 4

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

EASY

1th Jharkhand Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 8 - Binomial Theorem>EXERCISE 8.2>Q 6

Find the $13^{t h}$ term in the expansion of ${(9 x - \frac{1}{3 \sqrt{x}})}^{18}, x \neq 0$

MEDIUM

1th Jharkhand Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 8 - Binomial Theorem>EXERCISE 8.2>Q 7

Find the middle terms in the expansions of ${(3 - \frac{x^{3}}{6})}^{7}$ .

MEDIUM

1th Jharkhand Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 8 - Binomial Theorem>EXERCISE 8.2>Q 8

Find the middle term in the expansions of ${(\frac{x}{3} + 9 y)}^{10} .$

HARD

1th Jharkhand Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 8 - Binomial Theorem>EXERCISE 8.2>Q 9

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

HARD

1th Jharkhand Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 8 - Binomial Theorem>EXERCISE 8.2>Q 10

The coefficients of the ${(r - 1)}^{t h}, r^{t h}$ and ${(r + 1)}^{t h}$ terms in the expansion of ${(x + 1)}^{n}$ are in the ratio $1 : 3 : 5 .$ Find $n$ and $r .$

HARD

1th Jharkhand Board

IMPORTANT

MATHEMATICS Textbook for Class XI>Chapter 8 - Binomial Theorem>EXERCISE 8.2>Q 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 ${(1 + x)}^{2 n - 1} .$