The sum of the series ∑$$r=0n$$ $$($$−$$1)r$$ $$nCr12r+3r22r+7r23r+15r24r+$$…$$+m$$ terms is

Correct option is 12mn−12mn2n−1

Questions

The sum of the series $\sum_{r = 0}^{n} (- 1)^{r^{n}} C_{r} (\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \frac{15^{r}}{2^{4 r}} + \dots + m terms) is$

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2mn−12mn2n−1

2mn−12n−1

2mn+12n+1

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detailed solution

Correct option is A

∑nn (−1)rnCr12r+3r22r+7r23r+… upto m terms =∑r=0n (−1)r nCr⋅12r+∑r=0n (−1)r⋅nCr3r22r+∑r=0n (−1)rnCr7r23r+…=1−12n+1−34n+1−78n+…up to M terms=12n+122n+123n… upto m terms =12n1−12nm1−12n=2mn−12mn2n−1

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The sum of the series ∑r=0n (−1)r nCr12r+3r22r+7r23r+15r24r+…+m terms is

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The sum of the series $\sum_{r = 0}^{n} (- 1)^{r^{n}} C_{r} (\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \frac{15^{r}}{2^{4 r}} + \dots + m terms) is$