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

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(A) Let $\frac{2x+1}{(1+x)(1-2x)} = \frac{A}{1+x} + \frac{B}{1-2x} = \frac{A(1-2x) + B(1+x)}{(1+x)(1-2x)}$.$2x+1 = A(1-2x) + B(1+x)$.Comparing coeffic

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$\frac{5}{3}+\frac{8}{9}(4^5-1)$

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(A) Let $\frac{2x+1}{(1+x)(1-2x)} = \frac{A}{1+x} + \frac{B}{1-2x} = \frac{A(1-2x) + B(1+x)}{(1+x)(1-2x)}$.$2x+1 = A(1-2x) + B(1+x)$.Comparing coefficients,we get $A = -\frac{1}{3}$ and $B = \frac{4}{3}$.Thus,$\frac{2x+1}{(1+x)(1-2x)} = -\frac{1}{3}(1+x)^{-1} + \frac{4}{3}(1-2x)^{-1}$.$= -\frac{1}{3}[1-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+\dots] + \frac{4}{3}[1+2x+2^2x^2+2^3x^3+2^4x^4+2^5x^5+2^6x^6+2^7x^7+2^8x^8+2^9x^9+\dots]$.The coefficients of the first $5$ odd powers of $x$ (i.e.,$x^1, x^3, x^5, x^7, x^9$) are:For $x^1: -\frac{1}{3}(-1) + \frac{4}{3}(2) = \frac{1}{3} + \frac{8}{3} = 3$.Wait,calculating the sum directly from the expansion:Sum $= -\frac{1}{3}(-1-1-1-1-1) + \frac{4}{3}(2^1+2^3+2^5+2^7+2^9)$.Sum $= \frac{5}{3} + \frac{4}{3} 	imes \frac{2(4^5-1)}{4-1} = \frac{5}{3} + \frac{8}{9}(4^5-1)$.

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

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