Question

What is $\sum _{r=0}^1{}^{n+r}{C}_n$ equal to?

• A. ${}^{n+2}{C}_1$
• B. ${}^{n+2}{C}_n$
• C. ${}^{n+3}{C}_n$
• D. ${}^{n+2}{C}_{n-1}$

Answer 1

The correct option is A ${}^{n+2}{C}_1$

$\sum _{r=0}^1{}^{n+r}{C}_n={}^n{C}_n+{}^{n+1}{C}_n$

$=\frac{n!}{n!}+\frac{(n+1)!}{n!1!}$

$=1+n+1$

$=n+2$

$={}^{n+2}{C}_{n+1}={}^{n+2}{C}_1$