Question

If $C_r={}^{25}C_r$ and $C_0+5\cdot C_1+9\cdot C_2+\cdots +101·C_{25}=2^{25}·k$ then $k$ is equal to

Answer 1

Step 1. Find the value of $k$:

Let $S={}^{25}C_0+5{}^{25}C_1+9{}^{25}C_2+\dots .+97{}^{25}C_{24}+101{}^{25}C_{25}=2^{25}k$ …..(1)

Reverse the series and apply ${}^{\mathbf{n}}\mathit{C}_{\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}{}^{\mathbf{n}}\mathit{C}_{\mathbf{n}\mathbf{-}\mathbf{r}}$ in all coefficients,

$S=101{}^{25}C_0+97{}^{25}C_1+\dots +5{}^{25}C_{24}+{}^{25}C_{25}$ …..(2)

Step 2. Add equation (1) and (2),we get

$2S=102[{}^{25}C_0+{}^{25}C_1+\dots +{}^{25}C_5]$

$\Rightarrow$$S=51\times 2^{25}$

$\Rightarrow$$k=51$

Hence, The value of $\mathit{k}\mathbf{}\mathbf{=}\mathbf{}\mathbf{51}$