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

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Correct option is C

∑r=0k 22r7k kCr=17k∑r=0k kCr4r =17k(1+4)k=57kThus, ∑k=1∞ ∑r=0k 22r7k kCr =∑k=1∞ 57k=5/71−5/7=2.5

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Sum of the series ∑k=1∞ ∑r=0k 22r7k kCr is

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