If A = sum_{n=1}^{infty} frac{1}{(3+(-1)^{n}) | vedclass

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(C) For odd $n$,$3+(-1)^n = 3-1 = 2$. For even $n$,$3+(-1)^n = 3+1 = 4$.$A = \sum_{k=1}^{\infty} \frac{1}{2^{2k-1}} + \sum_{k=1}^{\infty} \frac{1}{4^{

Vedclass · Accepted Answer

$-\frac{11}{9}$

Vedclass · Answer

(C) For odd $n$,$3+(-1)^n = 3-1 = 2$. For even $n$,$3+(-1)^n = 3+1 = 4$.$A = \sum_{k=1}^{\infty} \frac{1}{2^{2k-1}} + \sum_{k=1}^{\infty} \frac{1}{4^{2k}} = (\frac{1}{2} + \frac{1}{2^3} + \dots) + (\frac{1}{4^2} + \frac{1}{4^4} + \dots)$$A = \frac{1/2}{1-1/4} + \frac{1/16}{1-1/16} = \frac{1/2}{3/4} + \frac{1/16}{15/16} = \frac{2}{3} + \frac{1}{15} = \frac{10+1}{15} = \frac{11}{15}$.$B = \sum_{k=1}^{\infty} \frac{-1}{2^{2k-1}} + \sum_{k=1}^{\infty} \frac{1}{4^{2k}} = (-\frac{1}{2} - \frac{1}{2^3} - \dots) + (\frac{1}{4^2} + \frac{1}{4^4} + \dots)$$B = \frac{-1/2}{1-1/4} + \frac{1/16}{1-1/16} = -\frac{2}{3} + \frac{1}{15} = \frac{-10+1}{15} = -\frac{9}{15} = -\frac{3}{5}$.$\frac{A}{B} = \frac{11/15}{-9/15} = -\frac{11}{9}$.

If $A = \sum_{n=1}^{\infty} \frac{1}{(3+(-1)^{n})^{n}}$ and $B = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(3+(-1)^{n})^{n}}$,then $\frac{A}{B}$ is equal to:

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