Question

The sum to (n+1) terms of the following series

$\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}$ + ......... is

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is D

None of these

$(1-x{)}^n=C_0-C_{1}x+C_2{x}^2-C_3{x}^3+.......$

⇒$x(1-x{)}^n=C_{0}x-C_1{x}^2+C_2{x}^3-C_3{x}^4+........$

⇒ ${\int }_0^{1}x(1-x{)}^{n}dx$ = ${\int }_0^1(C_{0}x-C_1{x}^2+C_2{x}^3.......)dx$ ..........(i)

The integral on the LHS

= ${\int }_1^0(1-t)t^n(-dt)$ , by putting 1 - x = t

= ${\int }_0^1(t^n-t^{n+1})dt$ = $\frac{1}{n+1}$ - $\frac{1}{n+2}$

Whereas the integral on the RHS of (i)

= [ $\frac{C_0{x}^2}{2}$ - $\frac{C_1{x}^3}{3}$ + $\frac{C_2{x}^4}{4}$ - ............] = $\frac{C_0}{2}$ - $\frac{C_1}{3}$ + $\frac{C_2}{4}$ - ........

∴ $\frac{C_0}{2}$ - $\frac{C_1}{3}$ + $\frac{C_2}{4}$ - .......... to (n+1) terms

= $\frac{1}{n+1}$ - $\frac{1}{n+2}$ = $\frac{1}{(n+1)(n+2)}$.