0

I'm currently working through a proof and I'm stuck at figuring out how $\sum_{i=1}^n ((\sum_{j=1}^k C_{ij})^2)$ becomes $\sum_{i=1}^n \sum_{j=1}^k C_{ij}^2 + 2\sum_{i=1}^n \sum_{j=1}^k \sum_{h=1, h \neq j}^k C_{ij}C_{ih}$

I've tried working through some basic summation algebra but I can't seem to find an identity or rule that gives me the second equation.

Cody R.
  • 1
  • 3
    It's just the binomial expansion of $(a+b+c+...)^2$. Each square term shows up once in the product, and each cross-product term shows up twice. – rogerl Mar 09 '21 at 22:44
  • Given rogerl's comment, I suggest setting $n=4$, $k=3$, setting up the two expressions using variables $C_{ij}$, and then manually comparing the results. – user2661923 Mar 09 '21 at 23:13
  • Thanks both of you! That helped clear things up. – Cody R. Mar 10 '21 at 01:51

0 Answers0