Question

${}^n{C}_0$ + 2.${}^n{C}_1$ + 3.${}^n{C}_2$ +..............(n+1)${}^n{C}_n$ =

• A. (n+1)
• B. (n+2)
• C. (n+2)
• D. (n+1)

Answer 1

The correct option is C

(n+2)

When we observe terms, each term is of the form (r+1)${}^n{C}_r$.

We can write the sum as

$\sum _{r=0}^n(r+1)$ ${}^n{C}_r$ = $\sum _{r=0}^n$ r ${}^n{C}_r$ + $\sum _{r=0}^n$ ${}^n{C}_r$

${}^n{C}_r$ = $\frac{n}{r}$${}^{n-1}{C}_{r-1}$

r ${}^n{C}_r$ = n ${}^{n-1}{C}_{r-1}$

⇒ $\sum _{r=0}^n(r+1)$ ${}^n{C}_r$ = $\sum _{r=0}^{n}r$${}^n{C}_r$ + $\sum _{r=0}^n$ ${}^n{C}_r$

= $\sum _{r=0}^n$ n × ${}^{n-1}{C}_{r-1}$ + $\sum _{r=0}^n$ ${}^n{C}_r$

= n × $\sum _{r=0}^n$ ${}^{n-1}{C}_{r-1}$ + $\sum _{r=0}^n$ ${}^n{C}_r$

$\sum _{r=0}^n$ ${}^n{C}_r$ = $2^n$, $\sum _{r=0}^n$ ${}^{n-1}{C}_{r-1}$ = $2^{n-1}$

⇒ $\sum _{r=0}^n(r+1)$ ${}^n{C}_r$ = n$2^{n-1}$ + $2^n$

= $2^{n-1}$ (n+2)