Double summation squared

Question

I need to calculate $(\sum_i \sum_tx_{i,t}-1)^2$. I know that $$ \sum_i \left(\sum_t x_{i,t}-1 \right)^2 = \sum_i\left(2\sum_t\sum_{u>t}x_{it}x_{iu} - \sum_tx_{it} +1\right) $$ from the common square of a sum, and normal binomial formula, but I'm not sure on how to approach adding the second sum into the square. I appreciate any help!

It is $$\left(\sum_i \sum_t x_{i,t}\right)^2-2\left(\sum_i \sum_t x_{i,t}\right)+1$$ which you can further expand — Maximilian Janisch, Apr 14 '20 at 22:03
But how to solve the first square in here? I only know how to square a single not a double sum — Robinbux, Apr 14 '20 at 22:11

score 1 · Accepted Answer · answered Apr 14 '20 at 22:13

1

One way is to use the shorthand $u_i = \sum_t x_{i,t}$ to get $$ \left(\sum_i \sum_t x_{i,t} - 1\right)^2 = \left(\sum_i u_i - 1\right)^2 $$ and apply the Binomial theorem to $u_i$ and then back-substitute.

answered Apr 14 '20 at 22:13

gt6989b

54,422

That's neat, I didn't of that. Thanks! – Robinbux Apr 14 '20 at 22:29

Double summation squared

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