In the expansion of (1+x)(1-x^2)(1+frac{3}{x} | vedclass

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(A) Given expression: $E = (1+x)(1-x^2)(1+\frac{1}{x})^{15} = (1+x)(1-x)(1+x)(1+\frac{1}{x})^{15}$ $= (1+x)^2(1-x) \cdot \frac{(x+1)^{15}}{x^{15}} = \

Vedclass · Accepted Answer

$118$

Vedclass · Answer

(A) Given expression: $E = (1+x)(1-x^2)(1+\frac{1}{x})^{15} = (1+x)(1-x)(1+x)(1+\frac{1}{x})^{15}$ $= (1+x)^2(1-x) \cdot \frac{(x+1)^{15}}{x^{15}} = \frac{(1-x^2)(1+x)^{16}}{x^{15}} = \frac{(1+x)^{16} - x^2(1+x)^{16}}{x^{15}}$ $= (1+x)^{16}x^{-15} - (1+x)^{16}x^{-13}$ Coefficient of $x^3$ in $E$: $= 	ext{coeff of } x^{18} 	ext{ in } (1+x)^{16} - 	ext{coeff of } x^{16} 	ext{ in } (1+x)^{16}$ $= 0 - \binom{16}{16} = -1$ Coefficient of $x^{-13}$ in $E$: $= 	ext{coeff of } x^2 	ext{ in } (1+x)^{16} - 	ext{coeff of } x^0 	ext{ in } (1+x)^{16}$ $= \binom{16}{2} - \binom{16}{0} = \frac{16 	imes 15}{2} - 1 = 120 - 1 = 119$ Sum of coefficients $= 119 + (-1) = 118$.

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

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