sum_{k=1}^{infty} sum_{r=0}^k frac{1}{3^k} bi | vedclass

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(D) We are given the expression $\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$.Using the identity $\sum_{r=0}^k \binom{k}{r} = 2^k$,we

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$2$

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(D) We are given the expression $\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$.Using the identity $\sum_{r=0}^k \binom{k}{r} = 2^k$,we can simplify the inner summation:$\sum_{k=1}^{\infty} \frac{1}{3^k} \left( \sum_{r=0}^k \binom{k}{r} \right) = \sum_{k=1}^{\infty} \frac{2^k}{3^k} = \sum_{k=1}^{\infty} \left( \frac{2}{3} \right)^k$.This is an infinite geometric series with the first term $a = \frac{2}{3}$ and common ratio $r = \frac{2}{3}$.The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.Substituting the values,we get $S = \frac{2/3}{1 - 2/3} = \frac{2/3}{1/3} = 2$.

$\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$ is equal to

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