If $C_{r} \equiv{ }^{25} C_{r}$ and $C_{0}+5 C_{1}+9 C_{2}+\ldots .+(101) \cdot C_{25}=2^{25} \cdot

Question

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

Professor PC Thomas Classes · Answer

(51) : We have, given sum\ $=\sum_{r=0}^{25}(4 r+1){ }^{25} C_{r}=4 \sum_{r=0}^{25} r \cdot{ }^{25} C_{r}+\sum_{r=0}^{25}{ }^{25} C_{r}$\ $=4 \sum_{r=1}^{25} r 	imes \frac{25}{r}{ }^{24} C_{r-1}+2^{25}=100 \sum_{r=1}^{25}{ }^{24} C_{r-1}+2^{25}$\ $=100 \cdot 2^{24}+2^{25}=2^{25}(50+1)=2^{25} \cdot 51$\ $=2^{25} \cdot k$ So, $k=51$