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 ____

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Binomial Theorem

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 coefficient of

x^{3}

and

x^{- 13}

is equal to ____

Answer: 118

Since

1 + \frac{3}{x} + \frac{3}{x ^{2}} + \frac{1}{x ^{3}} = (1 + \frac{1}{x})^{3}

, the expression is

(1 + x) (1 - x^{2}) (1 + \frac{1}{x})^{15} = (1 + x)^{2} (1 - x) \cdot \frac{( 1 + x ) ^{15}}{x ^{15}} = \frac{( 1 - x ) ( 1 + x ) ^{17}}{x ^{15}}

= \frac{( 1 + x ) ^{17}}{x ^{15}} - \frac{( 1 + x ) ^{17}}{x ^{14}}

Coefficient of

x^{3}

: this needs the coefficient of

x^{18}

(1 + x)^{17}

(which is

0

, since

17 < 18

) minus the coefficient of

x^{17}

(1 + x)^{17}

(which is

1

0 - 1 = - 1

.

Coefficient of

x^{- 13}

: needs the coefficient of

x^{2}

(1 + x)^{17}

, i.e.

(2 17) = 136

, minus the coefficient of

x^{1}

, i.e.

(1 17) = 17

136 - 17 = 119

.

Sum:

- 1 + 119 = 118

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Integer answer