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. $\frac{1}{n + 1}$
• B. $\frac{1}{n + 2}$
• C. $\frac{1}{n(n + 1)}$
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

$\mathbf{(1}\mathbf{-}\mathbf{x}\mathbf{)}^{\mathbf{n}}\mathbf{=}\mathbf{C}_{\mathbf{0}}\mathbf{-}\mathbf{C}_{\mathbf{1}}\mathbf{x +}\mathbf{C}_{\mathbf{2}}\mathbf{x}^{\mathbf{2}}\mathbf{-}\mathbf{C}_{\mathbf{3}}\mathbf{x}^{\mathbf{3}}\mathbf{+ .......}$

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

⇒ ${\int_{0}^{1}{x(1 - x)^{n}dx = C_{0}\left\lbrack \frac{x^{2}}{2} \right\rbrack_{0}^{1} - C_{1}\left\lbrack \frac{x^{3}}{3} \right\rbrack}}_{0}^{1} + C_{2}\left\lbrack \frac{x^{4}}{4} \right\rbrack_{0}^{1} - .......$ (i)

The integral on L.H.S. of (i) =$\int_{1}^{0}{(1 - t)t^{n}( - dt)}$ 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 R.H.S. of (i)

= $C_{0}\left\lbrack \frac{1}{2} \right\rbrack - C_{1}\left\lbrack \frac{1}{3} \right\rbrack + \frac{C_{2}}{4}.......$ = $\frac{C_{0}}{2} - \frac{C_{1}}{3} + \frac{C_{2}}{4} - .......$ to $(n + 1)$ terms = $\frac{1}{n + 1} - \frac{1}{n + 2} = \frac{1}{(n + 1)(n + 2)}$

Trick : Put $n = 1$ in given series = $\frac{1C_{0}}{2} - \frac{1C_{1}}{3} = \frac{1}{6}$.

Which is given by option (4).