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$^nC_1$ + $2.^nC_2$ + $3.^nC_3$ + ..... + $n.^nC_n$ = ?

$\begin{array}{1 1} n.2^n \\ n.2^{n-1} \\ (n+1).2^n \\ (n-1).2^{n} \end{array} $

1 Answer

$(1+x)^n=^nC_0+^nC_1x+^nC_1x^2+.......^nC_nx^n$
Differentiating both the sides w.r.t. $x$
$n(1+x)^{n-1}=^nC_1+2.^nC_2x+3.^nC_3x^2+......n.^nC_nx^{n-1}$
Put $x=1$ both the sides.
$n.2^{n-1}=^nC_1+^2.^nC_2+3.^nC_3+.......n.^nC_n$
answered Sep 24, 2013 by rvidyagovindarajan_1
 

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