If C_{0} + 5 cdot C_{1} + 9 cdot C_{2} + ldot | vedclass

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(C) Let $S = \sum_{r=0}^{25} (4r + 1) \cdot ^{25}C_{r}$.Using the property $^{n}C_{r} = ^{n}C_{n-r}$,we can write:$S = 1 \cdot ^{25}C_{0} + 5 \cdot ^{

Vedclass · Accepted Answer

$51$

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(C) Let $S = \sum_{r=0}^{25} (4r + 1) \cdot ^{25}C_{r}$.Using the property $^{n}C_{r} = ^{n}C_{n-r}$,we can write:$S = 1 \cdot ^{25}C_{0} + 5 \cdot ^{25}C_{1} + 9 \cdot ^{25}C_{2} + \ldots + 101 \cdot ^{25}C_{25}$.Reversing the order:$S = 101 \cdot ^{25}C_{25} + 97 \cdot ^{25}C_{24} + \ldots + 1 \cdot ^{25}C_{0}$.Adding the two expressions:$2S = \sum_{r=0}^{25} (4r + 1 + 101 - 4r) \cdot ^{25}C_{r} = \sum_{r=0}^{25} (102) \cdot ^{25}C_{r}$.$2S = 102 \sum_{r=0}^{25} {^{25}C_{r}} = 102 \cdot 2^{25}$.$S = 51 \cdot 2^{25}$.Comparing with $2^{25} \cdot k$,we get $k = 51$.

If $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \ldots + (101) \cdot C_{25} = 2^{25} \cdot k$,then $k$ is equal to:

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