Question

Find the sum ${\sum }_{r=1}^n\stackrel{n+r}{\underset{r}{C}}$.

Answer 1

${\sum }_{r=0}^n{}^{n+r}{C}_r={}^n{C}_0+{}^{n+1}{C}_1{+}^{n+2}{C}_2+..{.}^{n+n}{C}_n$

${\Rightarrow }^n{C}_n{+}^{n+1}{C}_n{+}^{n+2}Cn+..{.}^{2n}{C}_n$

$\Rightarrow$ Coefficient of $x^n$ in $\left[\right(1+x{)}^n+(1+x{)}^{n+1}+....+(1+x{)}^{2n}]$

$\Rightarrow$Coefficient of $x^n$ in $\frac{(1+x{)}^n[1-(1+x{)}^{n+1}]}{1-(1+x)}$

$\Rightarrow$Coefficient of $x^{n+1}$ in $\left[\right(1+x{)}^{2n+1}-(1+x{)}^n]$

${\Rightarrow }^{2n+1}{C}_{n+1}$

$\Rightarrow {\sum }_{r=0}^n{}^{n+r}{C}_r{=}^{2n+1}{C}_{n+1}$

$\Rightarrow {}^n{C}_0+{\sum }_{r=1}^n{}^{n+r}{C}_r{=}^{2n+1}{C}_{n+1}$

$\Rightarrow 1+{\sum }_{r=1}^n{}^{n+r}{C}_r{=}^{2n+1}{C}_{n+1}$

Therefore, $\Rightarrow {\sum }_{r=1}^n{}^{n+r}{C}_r{=}^{2n+1}{C}_{n+1}-1$