First slide

Binomial theorem for positive integral Index

Question

Sum of the series $\sum_{k = 1}^{\infty} \sum_{r = 0}^{k} \frac{2^{2 r}}{7^{k}} (^{k} C_{r})$ is

Moderate

A

$\frac{1}{7}$
B

$\frac{4}{7}$
C

$2.5$
D

$5.2$

Solution

$\begin{array}{l} \sum_{r = 0}^{k} \frac{2^{2 r}}{7^{k}} (^{k} C_{r}) = \frac{1}{7^{k}} \sum_{r = 0}^{k} (^{k} C_{r} 4^{r}) \\ = \frac{1}{7^{k}} (1 + 4)^{k} = {(\frac{5}{7})}^{k} \end{array}$
Thus, $\sum_{k = 1}^{\infty} \sum_{r = 0}^{k} \frac{2^{2 r}}{7^{k}} (^{k} C_{r})$
$= \sum_{k = 1}^{\infty} {(\frac{5}{7})}^{k} = \frac{5 / 7}{1 - 5 / 7} = 2.5$

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