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Having trouble figuring out this question. Any help would be appreciated! Thanks.

$\sum_{k=2}^n k(k-1)\binom{n}{k}$

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    $\begingroup$ What have you tried? Why don't you try a different form for $\binom{n}{k}$? The $k=2$ is a big hint. $\endgroup$ – michaelrccurtis Sep 7 '15 at 1:31
  • $\begingroup$ I simplified the original to $\sum_{k=2}^n \cfrac {n!}{(n-k!)/(k-2)!}$ which simplifies to $\sum_{k=2}^n \cfrac{n(n-1)*...*(n-k+1)}{(k-2)!}$ but do not know how to proceed. How does the k=2 factor in? Thank you. $\endgroup$ – Adam Taché Sep 7 '15 at 1:42
  • $\begingroup$ If you take your first simplification, can you think of a way to make that look like $\binom{n-2}{m}$ for some m, say? $\endgroup$ – michaelrccurtis Sep 7 '15 at 1:44
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First, we can substitute in the factorial form for the binomial coefficient and simplify to:

$$\sum_{k=2}^n \frac{n!}{(k-2)!(n-k)!}$$

If we then make the substitution $m = k-2$:

$$\sum_{m=0}^{n-2} \frac{n(n-1)(n-2)!}{m!((n-2)-m)!}$$

We can then bring out constant factors:

$$n(n-1)\sum_{m=0}^{n-2} \binom{n-2}{m}$$

Finally, we can note that the sum part is the expansion for $(1+1)^{n-2}$ (or from Pascals Triangle), which means that the result is:

$$n(n-1)2^{n-2}$$

Note that these types of simplification problem often appear, and the strategy is frequently to manipulate them into a form that looks like a binomial expansion of some kind.

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  • $\begingroup$ Well explained :-) $\endgroup$ – user220382 Sep 8 '15 at 7:18
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Consider the series $$S(t) = \sum_{k=0}^{n} \binom{n}{k} \, t^{k} = (1 + t)^{n}$$ then $$S''(t) = n(n-1) \, (1+t)^{n-2} = \sum_{k=0}^{n} k(k-1) \, \binom{n}{k} \, t^{k-2} = \sum_{k=2}^{n} k(k-1) \, \binom{n}{k} \, t^{k-2}$$ Now, by letting $t=1$ the identity $$\sum_{k=2}^{n} k(k-1) \, \binom{n}{k} = 2^{n-2} \, n(n-1)$$ is obtained.

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