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The sum to $(n + 1)$ terms of the series $\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - \frac{{{C_3}}}{5} + ...$ is
$\frac{1}{{n + 1}}$
$\frac{1}{{n + 2}}$
$\frac{1}{{n\left( {n + 1} \right)}}$
none of these
Solution
We have
${(1 – x)^n} = {C_0} – {C_1}x + {C_2}{x^2} – {C_3}{x^3} + …….$
$ \Rightarrow \quad x{(1 – x)^n} = {C_0}x – {C_1}{x^2} + {C_2}{x^3} – {C_3}{x^4} + …….$
$\int\limits_0^1 {x{{\left( {1 – x} \right)}^n}} dx = \int\limits_0^1 {x{{\left( {1 – x} \right)}^n}} dx$
$ = \int\limits_0^1 {\left( {1 – t} \right){t^n}} \left( { – 1} \right)dt\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[{\rm{ Put }}1 – x = t]$
$ = \int\limits_0^1 {\left( {{t^n} – {t^{n + 1}}} \right)} dt = \left| {\frac{{{t^{n + 1}}}}{{n + 1}} – \frac{{{t^{n + 2}}}}{{n + 2}}} \right|_0^1$
${=\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}}$
Integrating $R.H.S$ of $( 1)$ we get $\left. {\left( {\frac{{{C_0}{x^2}}}{2} – \frac{{{C_1}{x^3}}}{3} + \frac{{{C_2}{x^4}}}{4} + ……} \right)} \right|_0^1$
$=\frac{C_{0}}{2}-\frac{C_{1}}{3}+\frac{C_{2}}{4}-\ldots$
Thus $\quad \frac{C_{0}}{2}-\frac{C_{1}}{3}+\frac{C_{2}}{4}-\ldots=\frac{1}{(n+1)(n+2)}$
