Sum of the series ∑k=0n nCk(−1)k1akwhere ak=∑i=0k kCibiwhere bi=∑j=0i iCj−23jis

A
$\frac{1}{2^{n}}$
B
$\frac{1}{3^{n}}$
C
$\frac{1}{4^{n}}$
D
${(\frac{3}{4})}^{n}$

Solution:

$\begin{array}{l} b_{i} = \sum_{j = 0}^{i}^{i} C_{j} {(\frac{- 2}{3})}^{j} = {(1 - \frac{2}{3})}^{i} = {(\frac{1}{3})}^{i} \\ ∴ a_{k} = \sum_{i = 0}^{k}^{k} C_{i} {(\frac{1}{3})}^{i} = {(1 + \frac{1}{3})}^{k} = {(\frac{4}{3})}^{k} \end{array}$

Thus, $\sum_{k = 0}^{n}^{n} C_{k} (- 1)^{k} {(\frac{3}{4})}^{k} = {(1 - \frac{3}{4})}^{n} = \frac{1}{4^{n}}$

Sum of the series $\sum_{k = 0}^{n}^{n} C_{k} (- 1)^{k} \frac{1}{a_{k}}$
where $a_{k} = \sum_{i = 0}^{k}^{k} C_{i} b_{i}$
where $b_{i} = \sum_{j = 0}^{i}^{i} C_{j} {(\frac{- 2}{3})}^{j}$
is

Solution:

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