Sigma notation sequence

Question

How to calculate following

$\sum_{i<j=1}^n i*j$

I am not able to understand how to even write the nth term of this sequence. Please help in this.

Answer 1

I'll leave you to simplify$$\frac12\left(\left(\sum_ii\right)^2-\sum_ii^2\right)=\frac12\left(\frac14 n^2(n+1)^2-\frac16 n(n+1)(2n+1)\right).$$

Answer 2

Expanding on J.G.'s answer:

• J.G. is assuming the notation means "take the sum over all terms $i\cdot j$ where $1 \le i < j \le n$." For example, when $n=4$ this is $1 \cdot 2 + 1 \cdot 3 + 1 \cdot 4 + 2 \cdot 3 + 2 \cdot 4 + 3 \cdot 4$.
• What happens if you take $\sum_{i=1}^n i$ and square it? (E.g. $(1+2+3+4)^2 = 1^2 + 1 \cdot 2 + 1\cdot 3 + \cdots + 3 \cdot 4 + 4^2$.) First, you get things like $1^2$, $2^2$, $3^2$, which you don't want. After subtracting those terms away, you have twice as many terms as you need (you want terms like $2 \cdot 4$ where the first factor is smaller than the second, but for each of these you have a corresponding term $4 \cdot 2$). So J.G. starts with $\left(\sum_{i=1}^n i\right)^2$, subtracts away the perfect squares $\sum_{i=1}^n i^2$, and divides by two.
• Useful facts: $\sum_{i=1}^n i = \frac{n(n+1)}{2}$ and $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$.