In the expansion of ( 1 + x ) ( 1 - x 2 ) ( 1 + 3 x + 3 x 2 + 1 x 3 ) 5 , x ? 0 , the sum of the coefficients of x 3 and x - 13 is equal to _______ . [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
In the expansion of $(1 + x) (1 - x^{2}) {(1 + \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}})}^{5}, x \neq 0,$ the sum of the coefficients of $x^{3}$ and $x^{- 13}$ is equal to _______ . [2024]

Ans.
(118)

The given expansion $(1 + x) (1 - x^{2}) {(1 + \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}})}^{5}$

$\Rightarrow {(1 + x)}^{2} (1 - x) \frac{{(x^{3} + 3 x^{2} + 3 x + 1)}^{5}}{x^{15}}$

$\Rightarrow \frac{{(1 + x)}^{2} (1 - x) [{(x + 1)}^{3}]^{5}}{x^{15}} \Rightarrow \frac{(1 - x) {(1 + x)}^{17}}{x^{15}}$

$∴$ Coefficient of $x^{3} \Rightarrow x^{18} in (1 - x) {(1 + x)}^{17}$

$\Rightarrow {(1 + x)}^{17} - x {(1 + x)}^{17}$

$\Rightarrow 0 - x ({}^{17}{C_{17}} x^{17}) = {}^{17}{C_{17}} (- 1) = - 1$

and coefficient of $x^{- 13} \Rightarrow x^{2} in (1 - x) {(1 + x)}^{17}$

$\Rightarrow {(1 + x)}^{17} - x {(1 + x)}^{17}$

$\Rightarrow {}^{17}{C_{2}} - {}^{17}{C_{1}} = 17 \times 8 - 17 = 17 \times 7 = 119$

The sum of the coefficient $x^{3}$ and $x^{- 13} = - 1 + 119 = 118$

Want Us to Email you About Special Offers & Updates?