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$^nC_0$ - $\large\frac{^nC_1}{2}$ + $\frac{^nC_2}{3}$ - ...... = ?

$\begin{array}{1 1} 0 \\ 2^{n-1} \\ \frac{1}{n+1}\\ 2^{n+1} \end{array} $

1 Answer

$(1-x)^n=^nC_0-^nC_1x+^nC_2x^2-..............$
Integrate both the sides w.r.t. $x$.
$-\large\frac{1}{n+1}(1-x)^{n+1}$$+c=^nC_0x-\large\frac{^nC_1.x^2}{2}+\frac{^nC_2 x^3}{3}-...........$
Put $x=0$, then $c=\large\frac{1}{n+1}$
Then put $x=1$, we get
$^nC_0-\large\frac{^nC_1}{2}+\frac{^nC_2}{3}-.......=\frac{1}{n+1}$
answered Sep 26, 2013 by rvidyagovindarajan_1
 
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