I am trying to evaluate the sum of binomial coefficients multiplied by the terms of an arithmetic series. In my particular case, I am hoping to evaluate
$$ \sum_{i=0}^{2n}(6n – 3i){3n\choose i}.$$
Is a closed form expression for this sum available?
Many thanks in advance!
Best Answer
Following from the suggestion above, we can rewrite: $$\sum_{k=0}^{2m}(6m-3k)\cdot {}_{3m} \mathrm{C}_k$$ Factoring and separating, $$3\left(\sum_{k=0}^{2m}2m\cdot {}_{3m} \mathrm{C}_k - \sum_{k=0}^{2m}k\cdot {}_{3m} \mathrm{C}_k\right)$$ Focusing on the first sum, $$2m\sum_{k=0}^{2m} {}_{3m} \mathrm{C}_k$$ $$= 2m\frac{(3m)!}{(2m)!}\sum_{k=0}^{2m} {}_{2m} \mathrm{C}_k$$ And using the important identity $$\sum_{k=0}^{N} {}_N \mathrm{C}_k=2^N$$ $$2m\frac{(3m)!}{(2m)!}\sum_{k=0}^{2m} {}_{2m} \mathrm{C}_k=\frac{(3m)!}{(2m-1)!} 2^{2m}$$ Now onto the other sum. $$\sum_{k=0}^{2m}k\cdot {}_{3m} \mathrm{C}_k=\frac{(3m)!}{(2m)!}\sum_{k=0}^{2m}k\cdot {}_{2m} \mathrm{C}_k$$ Using the results found from this question we can evaluate this to be $$\frac{(3m)!}{(2m)!}2^{2m-1}\cdot 2m= \frac{(3m)!}{(2m-1)!}2^{2m-1}$$ Thus, $$3\left(\sum_{k=0}^{2m}2m\cdot {}_{3m} \mathrm{C}_k - \sum_{k=0}^{2m}k\cdot {}_{3m} \mathrm{C}_k\right)$$ $$=3\left(\frac{(3m)!}{(2m-1)!} 2^{2m} - \frac{(3m)!}{(2m-1)!}2^{2m-1}\right)$$ $$=\frac{3(3m)!}{(2m-1)!}2^{2m-1}$$ Now using $n=3m$ (which was edited out by OP), $$=\frac{2n}{3} \frac{3\cdot n!}{(\frac{2n}{3}-1)!}2^{\frac{2n}{3}-1}$$ $$=n \frac{n!}{(\frac{2n}{3}-1)!}2^{\frac{2n}{3}}.$$