I'm working on proofs, and I came across the concepts generating functions and Burnside's lemma. So the summation is this: $\sum_{k=0}^{n-1}(2n-2k){n \choose k} = n2^{n} $. (If someone could format this that would be great).
How would you proove this summation using generating functions? How is Burnside's lemma used to solve this problem also?
On
Starting with $$n \, (1+ t)^{n-1} = \frac{d}{dt} (1 + t)^{n} = \frac{d}{dt} \, \sum_{k=0}^{n} \binom{n}{k} \, t^{k} = \sum_{k=0}^{n} \binom{n}{k} \, k \, t^{k-1}$$ then $$2^{n} \, n = 2 \, \sum_{k=0}^{n} \binom{n}{k} \, k.$$ Now, $$\sum_{k=0}^{n} \binom{n}{k} \, k = \sum_{k=0}^{n} \binom{n}{k} \, (n-k)$$ and leads to $$\sum_{k=0}^{n} \binom{n}{k} \, (2 n - 2k) = 2^{n} \, n.$$
Hint:
$\sum_{k=0}^{n-1}(2n-2k){n \choose k} =2\sum_{k=0}^{n-1}(n-k){n \choose k} =2\sum_{k=1}^{n}k{n \choose k} $.