sumlimits_{n = 1}^infty {sumlimits_{k = 1}^{ | vedclass

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(B) माना $S = \sum\limits_{n = 1}^\infty \sum\limits_{k = 1}^{n - 1} \frac{k}{2^{n+k}} = \sum\limits_{n = 1}^\infty \frac{1}{2^n} \sum\limits_{k = 1}^

Vedclass · Accepted Answer

$\frac{4}{9}$

Vedclass · Answer

(B) माना $S = \sum\limits_{n = 1}^\infty \sum\limits_{k = 1}^{n - 1} \frac{k}{2^{n+k}} = \sum\limits_{n = 1}^\infty \frac{1}{2^n} \sum\limits_{k = 1}^{n - 1} \frac{k}{2^k}$.हम जानते हैं कि $\sum\limits_{k = 1}^{n - 1} k x^k = \frac{x(1 - nx^{n-1} + (n-1)x^n)}{(1-x)^2}$.$x = \frac{1}{2}$ के लिए,$\sum\limits_{k = 1}^{n - 1} \frac{k}{2^k} = 2 - \frac{n+1}{2^{n-1}}$.इसे प्रतिस्थापित करने पर,$S = \sum\limits_{n = 1}^\infty \frac{1}{2^n} (2 - \frac{n+1}{2^{n-1}}) = \sum\limits_{n = 1}^\infty (\frac{1}{2^{n-1}} - \frac{n+1}{2^{2n-1}})$.$S = 2 - 2 \sum\limits_{n = 1}^\infty \frac{n+1}{4^n} = 2 - 2(\frac{7}{9}) = \frac{4}{9}$.

$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^{n - 1} {\frac{k}{{{2^{n + k}}}}} } $ का मान ज्ञात कीजिए।

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