The sum of the series ∑r=0n(−1)r.ncr[12r+3r22r+7r23r+15r24r+...m terms] is equal to

S= k=1m ∑r=0m(−1)n(ncr)(2k−12r)n= k=1m(1−2k−12k)n= k=1m(12n)k=12n+(12n)2+.....+(12n)m= 12n(1−12mn)|1−12n|=2mn−12mn(2n−1)

The sum of the series ∑r=0n(−1)r.ncr[12r+3r22r+7r23r+15r24r+...m terms] is equal to

S= k=1m ∑r=0m(−1)n(ncr)(2k−12r)n= k=1m(1−2k−12k)n= k=1m(12n)k=12n+(12n)2+.....+(12n)m= 12n(1−12mn)|1−12n|=2mn−12mn(2n−1)

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The sum of the series $\sum_{r = 0}^{n} {(- 1)}^{r} .^{n} c_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \frac{15^{r}}{2^{4 r}} + ... m t e r m s]$ is equal to

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

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

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

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

answer is A.

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Detailed Solution

$S =_{k = 1}^{m} \sum_{r = 0}^{m} {(- 1)}^{n} (^{n} c_{r}) {(\frac{2^{k} - 1}{2^{r}})}^{n} =_{k = 1}^{m} {(1 - \frac{2^{k} - 1}{2^{k}})}^{n} =_{k = 1}^{m} {(\frac{1}{2^{n}})}^{k}$
$= \frac{1}{2^{n}} + {(\frac{1}{2^{n}})}^{2} + ..... + {(\frac{1}{2^{n}})}^{m} = \frac{\frac{1}{2^{n}} (1 - \frac{1}{2^{m n}})}{| 1 - \frac{1}{2^{n}} |} = \frac{2^{m n} - 1}{2^{m n} (2^{n} - 1)}$