Question

Find ${}^n{C}_0$ - 2${}^n{C}_1$ + 3${}^n{C}_2$ - 4${}^n{C}_3$ .....................+ $(-1{)}^n$ (n+1) ${}^n{C}_n$

• A. 0
• B. (n+2)
• C. (n+1)
• D. (n+1)

Answer 1

The correct option is A

0

Each term in this sum is of the form $(-1{)}^r(r+1)$ × ${}^n{C}_r$.

So the sum can be co-written as

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

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

We know ${}^n{C}_r$ = $\frac{n}{r}$ ${}^{n-1}{C}_{r-1}$, when r > 0

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

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

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

Both of them will add upto zero [ First term is the sum of the coefficients of $(1-x{)}^{n-1}$ and 2nd term is

the sum of the coefficients of $(1-x{)}^n$]