Mathematics - Binomial Theorem Question with Solution | TestHub

$r\cdot \frac{{\left({}^n{C}_r\right)}^2}{{}^n{C}_{r-1}}=(n-r+1)c_r$ $\begin{array}{rlr}& \sum _{r=1}^{n}r\cdot \frac{c_r^2}{c_{r-1}}=\sum _{r=1}^n\{(n+1)-r\}{c}_r=(n+1)\left(2^n-1\right)-n{2}^{n-1}& =(n+2)2^{n-1}-(n+1)\end{array}$ $\mathrm{S}=c_0^2+{\sum }_{r=1}^{n}r\cdot \frac{c_r^2}{c_{r-1}}=(n+2)2^{n-1}-n$ $\mathrm{S}+\mathrm{n}=(n+2)2^{n-1}$ which is divisible by $(\mathrm{n}+2)$ and $2^{n-1}$ also, s is a positive integer.