The sum of the series sumlimits_{r = 0}^n {(- | vedclass

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(A) The given series is $\sum\limits_{r = 0}^n {(-1)^r \, ^nC_r \left( \sum\limits_{k = 1}^m \frac{(2^k - 1)^r}{2^{kr}} \right)}$.By interchanging the

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$\frac{2^{mn} - 1}{2^{mn}(2^n - 1)}$

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(A) The given series is $\sum\limits_{r = 0}^n {(-1)^r \, ^nC_r \left( \sum\limits_{k = 1}^m \frac{(2^k - 1)^r}{2^{kr}} \right)}$.By interchanging the order of summation,we get:$\sum\limits_{k = 1}^m \sum\limits_{r = 0}^n {^nC_r (-1)^r \left( \frac{2^k - 1}{2^k} \right)^r}$.Using the binomial theorem $\sum\limits_{r=0}^n {^nC_r x^r = (1+x)^n}$,we have:$= \sum\limits_{k = 1}^m \left( 1 - \frac{2^k - 1}{2^k} \right)^n$.$= \sum\limits_{k = 1}^m \left( \frac{2^k - 2^k + 1}{2^k} \right)^n = \sum\limits_{k = 1}^m \left( \frac{1}{2^k} \right)^n = \sum\limits_{k = 1}^m \frac{1}{2^{nk}}$.This is a geometric progression with $m$ terms,first term $a = \frac{1}{2^n}$ and common ratio $r = \frac{1}{2^n}$.The sum is $S_m = a \frac{1 - r^m}{1 - r} = \frac{1}{2^n} \frac{1 - (1/2^n)^m}{1 - 1/2^n} = \frac{1}{2^n} \frac{(2^{nm} - 1)/2^{nm}}{(2^n - 1)/2^n} = \frac{2^{nm} - 1}{2^{nm}(2^n - 1)}$.

The sum of the series $\sum\limits_{r = 0}^n {(-1)^r \, ^nC_r \left( \frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \frac{15^r}{2^{4r}} + \dots + m \text{ terms} \right)}$ is

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