double summation problem $\sum^5_{i=1}i \times \sum^5_{j=1}j =....$ please check

Question

(I) $\sum^5_{i=1}i \times \sum^5_{j=1}j = 1 \times (1) +1 \times (2) + \cdots +1\times (5) +2\times (1)+2\times (2) +\cdots + 2\times (5) + 3\times (1) + 3\times (2) + \cdots +3\times (5) + 4\times (1) +\cdots +4\times (5) + 5\times (1) + \cdots 5\times (5)$

(II) $\sum\sum^5_{0<i<j\leq 5} ij = 1 \times (2) + 1 \times (3) \cdots +1\times (5) +2\times (3) +\cdots + 2\times (5) + 3\times (4) + 3\times (5) + 4\times (5) $

Is the above summation correct please suggest if there is any correction required thanks.

Answer 1

The first sum is correct.

The second sum can be rewritten as follows:

$\displaystyle \sum_{i=1}^4 \sum_{j=i+1}^5 ij$, and it gives the same result as yours.

Answer 2

1. For the first sum, as @Ricardo said, you will have $$\sum^5_{i=1}i \times \sum^5_{j=1}j=\sum^5_{i=1}i \times \sum^5_{i=1}i=\left(\sum^5_{i=1}i\right)^2.$$

2. For the second sum, $$ \sum^5_{0<i<j\leq 5} ij=\sum_{i=1}^5 \sum_{j>i}^5 ij=\frac12\sum_{i=1}^5 \sum_{j\neq i}^5 ij=\frac12\sum_{i=1}^5 \sum_{j=1}^5 ij-\frac12\sum_{i=1}^5 i^2=\frac12\left\{\left(\sum^5_{i=1}i\right)^2-\sum_{i=1}^5 i^2\right\}. $$