Visual SolutionPYQ 2024 · Jan 31 Shift 1Tricky

Question

In the expansion of

(1+x)(1-x^2)\left(1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3}\right)^5

x \neq 0

, the sum of the coefficient of

x^3

and

x^{-13}

is equal to _____.

Solution Path

Simplify to

\frac{(1+x)^{17}(1-x)}{x^{15}}

. Coeff of

x^3

-1

. Coeff of

x^{-13}

136 - 17 = 119

. Sum

= 118

01Question Setup

1/4

Find the sum of coefficients of

x^3

and

x^{-13}

(1+x)(1-x^2)(1 + 3/x + 3/x^2 + 1/x^3)^5

[x^3] + [x^{-13}] = \;?

02Simplify

2/4

Recognize

(1 + 1/x)^3

pattern. Expression simplifies to

\frac{(1+x)^{17}(1-x)}{x^{15}}

\frac{(1+x)^{17}(1-x)}{x^{15}}

03Extract CoefficientsKEY INSIGHT

3/4

[x^3]

: need

[x^{18}]

(1+x)^{17}(1-x) = -1

[x^{-13}]

: need

[x^2] = {}^{17}C_2 - {}^{17}C_1 = 119

-1 + 119 = 118

04Final Answer

4/4

Sum of coefficients is

118

\boxed{118}

Concepts from this question1 concepts unlocked

★

EASY

T(r+1) = C(n,r) * a^(n-r) * b^r gives any specific term without expanding fully

T_{r+1} = {}^nC_r\, a^{n-r}\, b^r

Core formula for finding specific terms, term independent of x, rational terms, etc. Used in 4-5 questions every JEE paper.

Specific termIndependent termRational/integral termsCoefficient finding

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