The sum to (n + 1) terms of the series frac{{ | vedclass

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Question

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} -

Vedclass · Accepted Answer

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Vedclass · Answer

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 $\quad \frac{C_{0}}{2}-\frac{C_{1}}{3}+\frac{C_{2}}{4}-\ldots=\frac{1}{(n+1)(n+2)}$

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

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