If the coefficient of x^{13} in the expansion | vedclass

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(C) The expression is $\frac{(1+x)^2}{(1-2x)^3} = (1+2x+x^2)(1-2x)^{-3}$.Using the binomial expansion $(1-z)^{-n} = \sum_{r=0}^{\infty} \binom{n+r-1}{

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$1724$

Vedclass · Answer

(C) The expression is $\frac{(1+x)^2}{(1-2x)^3} = (1+2x+x^2)(1-2x)^{-3}$.Using the binomial expansion $(1-z)^{-n} = \sum_{r=0}^{\infty} \binom{n+r-1}{r} z^r$,we have $(1-2x)^{-3} = \sum_{r=0}^{\infty} \binom{r+2}{2} (2x)^r$.The coefficient of $x^r$ in $(1-2x)^{-3}$ is $\binom{r+2}{2} 2^r$.We need the coefficient of $x^{13}$ in $(1+2x+x^2) \sum_{r=0}^{\infty} \binom{r+2}{2} 2^r x^r$.This is equal to $1 \cdot \binom{13+2}{2} 2^{13} + 2 \cdot \binom{12+2}{2} 2^{12} + 1 \cdot \binom{11+2}{2} 2^{11}$.$= \binom{15}{2} 2^{13} + 2 \cdot \binom{14}{2} 2^{12} + \binom{13}{2} 2^{11}$.$= 105 \cdot 2^{13} + 91 \cdot 2^{13} + 78 \cdot 2^{11}$.$= 2^{11} (105 \cdot 4 + 91 \cdot 4 + 78) = 2^{11} (420 + 364 + 78) = 2^{11} (862) = 2^{10} (1724)$.Thus,$A = 1724$.

If the coefficient of $x^{13}$ in the expansion of $\frac{(1+x)^2}{(1-2x)^3}$ is $A \times 2^{10}$,then $A=$

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