Statement -1: _r=0^n(r+1),^nC_r=(n+2)2^n-1 Statement-2: _r=0^n(r+1),^nC_rx^r=(1+x)^n+nx(1+…

Question

Statement -1: $\displaystyle\sum_{r=0}^{n}(r+1)\,{}^nC_r=(n+2)2^{n-1}$

Statement-2: $\displaystyle\sum_{r=0}^{n}(r+1)\,{}^nC_rx^r=(1+x)^n+nx(1+x)^{n-1}$

Accepted Answer

$$\sum_{r=0}^n(r+1){}^nC_rx^r = \sum_{r=0}^n r\,{}^nC_rx^r+\sum_{r=0}^n{}^nC_rx^r = nx\sum_{r=1}^n{}^{n-1}C_{r-1}x^{r-1}+(1+x)^n = nx(1+x)^{n-1}+(1+x)^n$$

Statement 2 is true. Putting $x=1$:

$$\sum_{r=0}^n(r+1){}^nC_r = n\times2^{n-1}+2^n=2^{n-1}(n+2)$$

Statement 1 is also true, and Statement 2 is a correct explanation for it.

Solutions · 1