Why is $ \sum_{k = 1}^{n - 1}O\left( \binom{n}{k}k^2 \right) = O(n^22^n)$? In the analysis of an exact dynamic programming analysis for the Travelling Salesman problem in Exact Exponential Algorithms by Fomin and Kratsch, it is stated on p. 6 that
$$ \sum_{k = 1}^{n - 1}O\left( \binom{n}{k}k^2 \right) = O(n^22^n)$$
I would like some help to see why this is true. I understand that 
$$ \sum_{k = 1}^{n - 1} \binom{n}{k} = O(2^n) \qquad \text{and} \qquad \sum_{k = 1}^{n - 1} k^2 = O(n^3) $$
I figure the former in particular is relevant, but I can't see how multiplying by the $k^2$ term inside the sum gets the $n^2$ term in the right-hand side.

The Big - O notation is additive, so you just need to prove:
$$\sum_{k=1}^{n-1}\binom{n}{k}k^2 = O(n^22^n).$$
Note that you don't even need to explicitly calculate the sum, though it is possible. 
Here is a start: 
$$\sum_{k=1}^{n-1}\binom{n}{k}k^2\leq \sum_{k=1}^{n-1}\binom{n}{k}n^2=...$$
