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$\sum ^{n}_{r=0} r^2.\:^nC_r$ = ?

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

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Toolbox:
  • $\sum ^{n}_{r=0}^nC_r=2^n$
  • $\sum ^{n}_{r=1} ^{n-1}C_{r-1}=2^{n-1}$
  • $\sum ^{n}_{r=2} ^{n-2}C_{r-2}=2^{n-2}$
  • $^nC_r=\large\frac{n}{r}$ $.^{n-1}C_{r-1}=\large\frac{n(n-1)}{r(r-1)}$. $^{n-2}C_{r-2}$
$^nC_r=\large\frac{n(n-1)}{r(r-1)}$. $^{n-2}C_{r-2}$
$r^2.\:^nC_r=(r^2-r+r).^nC_r$
$=(r^2-r).^nC_r+r.^nC_r$
$=r(r-1).^nC_r+r.^nC_r$
$=r(r-1).\large\frac{n(n-1)}{r(r-1)}$$.^{n-2}C_{r-2}+r.\large\frac{n}{r}$ $.^{n-1}C_{r-1}$
$=n(n-1).^{n-2}C_{r-2}+n.^{n-1}C_{r-1}$
$\therefore\:\:\sum ^{n}_{r=0}\:r^2.\:^nC_r=\sum^{n}_{r=2} n(n-1).^{n-2}C_{r-2}+\sum^{n}_{r=1}n.^{n-1}C_{r-1}$
$=n(n-1).2^{n-2}+n.2^{n-1}$
$=2^{n-2}.(n^2-n+2n)=n(n+1).2^{n-2}$
answered Oct 22, 2013 by rvidyagovindarajan_1
 

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