Having trouble figuring out this question. Any help would be appreciated! Thanks.
$\sum_{k=2}^n k(k-1)\binom{n}{k}$
Answer
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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