If |x|

Question

(D) We have the expression $\frac{1+2x}{(1-2x)^2} = (1+2x)(1-2x)^{-2}$.Using the binomial expansion for negative indices,$(1-y)^{-2} = \sum_{k=0}^{\in

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$(2r+1) 2^r$

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(D) We have the expression $\frac{1+2x}{(1-2x)^2} = (1+2x)(1-2x)^{-2}$.Using the binomial expansion for negative indices,$(1-y)^{-2} = \sum_{k=0}^{\infty} (k+1)y^k$.Substituting $y = 2x$,we get $(1-2x)^{-2} = \sum_{k=0}^{\infty} (k+1)(2x)^k = \sum_{k=0}^{\infty} (k+1) 2^k x^k$.Now,multiply by $(1+2x)$:$(1+2x) \sum_{k=0}^{\infty} (k+1) 2^k x^k = \sum_{k=0}^{\infty} (k+1) 2^k x^k + \sum_{k=0}^{\infty} 2(k+1) 2^k x^{k+1}$.To find the coefficient of $x^r$,we take the term where $k=r$ from the first sum and the term where $k+1=r$ (i.e.,$k=r-1$) from the second sum:Coefficient of $x^r = (r+1) 2^r + 2((r-1)+1) 2^{r-1}$.$= (r+1) 2^r + 2(r) 2^{r-1} = (r+1) 2^r + r 2^r$.$= (r+1+r) 2^r = (2r+1) 2^r$.

If $|x| < \frac{1}{2}$,then the coefficient of $x^r$ in the expansion of $\frac{1+2x}{(1-2x)^2}$ is

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