Question

Find $\frac{1}{2}$${}^n{C}_1$ - $\frac{2}{3}$${}^n{C}_2$ + $\frac{3}{4}$${}^n{C}_3$.................$\frac{(-1{)}^{n+1}n}{n+1}$×${}^n{C}_n$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Each term is of the form .$\frac{(-1{)}^{r+1}r}{r+1}$×${}^n{C}_r$, where r varies from 1 to n.

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

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

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

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

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

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

= (${}^n{C}_1$ - ${}^n{C}_2$ + ${}^n{C}_3$..........$(-1{)}^{n+1}$ ${}^n{C}_n$)

- $\frac{1}{n+1}$(${}^{n+1}{C}_2$ - ${}^{n+1}{C}_3$ ...............$(-1{)}^{n+1}$ ${}^{n+1}{C}_{n+1}$)

= [(${}^{-}n{C}_0$ + ${}^n{C}_1$ - ${}^n{C}_2$...............) + ${}^n{C}_0$]