In the expansion of $\frac{2 x+1}{(1+x)(1-2 x)}$, the sum of the coefficients of the first 5 odd powers of $x$ is

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$\begin{aligned} & 	ext { Let } \frac{2 x+1}{(1+x)(1-2 x)} \ & =\frac{A}{1+x}+\frac{B}{1-2 x}=\frac{A(1-2 x)+B(1+x)}{(1+x)(1-2 x)} \ & \Rightarrow 2 x+1=A(1-2 x)+B(1+x)\end{aligned}$ On comparing both sides, we get $\begin{aligned} & A=-\frac{1}{3} 	ext { and } B=\frac{4}{3} \ & 	herefore \frac{2 x+1}{(1+x)(1-2 x)}=\frac{-1}{3}(1+x)^{-1}+\frac{4}{3}(1-2 x)^{-1} \ & =\frac{-1}{3}\left[1-x+x^2-\ldots \inftyight]+\frac{4}{3}\left[1+2 x+2^2 x^2+2^3 x^3+\ldots \inftyight] \end{aligned}$ Sum of the coefficient of the first 5 odd power of $x$ $\begin{aligned} & =\frac{-1}{3}[-1-1-1-1-1]+\frac{4}{3}\left[2+2^3+2^5+2^7+2^9ight] \ & =\frac{5}{3}+\frac{4}{3} \cdot\left[\frac{2\left(4^5-1ight)}{4-1}ight]=\frac{5}{3}+\frac{8}{9}\left(4^5-1ight) \end{aligned}$

In the expansion of $\frac{2 x+1}{(1+x)(1-2 x)}$, the sum of the coefficients of the first 5 odd powers of…

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