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$\sum^{n} _ {r=0} r. \: ^nCr$ = ?

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

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  • $^nC_r=\large\frac{n}{r}.$$^{n-1}C_{r-1}$
  • $\sum ^{n}_{r=0}^nC_r=2^n$
  • $\sum ^{n}_{r=1}^{n-1}C_{r-1}=2^{n-1}$
\[\sum ^{n}_{r=0}r.\:\:^nC_r=\sum ^{n}_{r=1}r.\large\frac{n}{r}.\:\: ^{n-1}C_{r-1}\]
\[=\sum ^{n} _{r=1} n.\:\:^{n-1}C_{r-1}\]
\[=n.\:\sum ^{n} _{r=1} \:\: ^{n-1}C_{r-1}\]
$=n.2^{n-1}$
answered Oct 22, 2013 by rvidyagovindarajan_1
edited Oct 22, 2013 by rvidyagovindarajan_1
 

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