I found this problem that I'm not sure how to solve. I would appreciate if anyone could point me to the right direction.
I need to find the following sum:
$\sum_{i+j+k=7} (-1)^i(-1)^j\frac{7!}{i!j!k!}$ where $i,j,k$ are elements of $\mathbb{N_0}$
There must be a much better way to solve this other than writing down all possible combinations and sum them (please correct me if I'm wrong).
The thing I believed could be useful was the multinomial theorem (I apologize if this is not the correct term in English), which states that:
$(\sum_{i=1}^k x_i)^n=\sum_{\sum_{i=1}^k n_i =n}\frac{n!}{\prod_{i=1}^k n_i!}\prod_{i=1}^k x_i^{n_i}$
But I'm missing $(-1)^k$.
I'm a bit new to this subject, but couldn't find any example like this in my books. Thank you in advance!
EDIT: My apologies to anyone who read the original question, I've misplaced $i$ and $j$, didn't see it while revising what I wrote. It's fixed now.
Answer
There is no $(-1)^k$ because $x_3$ is going to be $1$: $$-1=((-1)+(-1)+1)^7=\sum_{i+j+k=7} (-1)^i(-1)^j(1)^k\frac{7!}{i!j!k!}$$
No comments:
Post a Comment