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Sum of $(n+1)$ terms of the series $\large\frac{^nC_0}{2}$ - $\frac{^nC_1}{3}$ + $\frac{^nC_2}{4}$ - ...... = ?

$\begin{array}{1 1} (A) \large\frac{1}{n+1} \\ (B) \large\frac{1}{n+2} \\ (C) \large\frac{1}{n(n+1)} \\ (D) \large\frac{1}{(n+1)(n+2)} \end{array} $

1 Answer

$(1-x)^n=^nC_0-^nC_1x+^nC_2x^2-.............$
$x(1-x)^n=^nC_0x-^nC_1x^2+^nC_2x^3-..........$
$\Rightarrow\:\int ^1_0 x(1-x)\,dx=\int ^1 _0 (^nC_0x-^nC_1x^2+^nC_2x^3-........)\,dx$
$\Rightarrow\:\large\frac{1}{(n+1)(n+2)}=\bigg[\frac{^nC_0x^2}{2}-\frac{^nC_1x^3}{3}+\frac{^nC_2x^4}{4}-......\bigg]\bigg|^1 _0$
$\Rightarrow\:\large\frac{^nC_0}{2}-\frac{^nC_1}{3}+\frac{^nC_2}{4}-......=\frac{1}{(n+1)(n+2)}$
answered Oct 16, 2013 by rvidyagovindarajan_1
 
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