If $q_i = q_j$ for all $i \neq j$ what's $\sum_{i=1}^n{\sum_{j=1}^{i-1}{q_i q_j}}$? (need confirmation)

Question

I need to evaluate the following sum and wanna recheck here that I'm not mistaken. So can you please verify the corrctness or hint me to a mistake \begin{align} &\sum_{i=1}^n{\sum_{j=1}^{i-1}{q_i q_j}}\\ = &q_1q_2 + \ldots + q_1 q_n + q_2q_3 + \ldots q_2q_n + \ldots + q_{n-1}q_n \\ \stackrel{q_i = q}{=} &\underbrace{q^2 + \ldots + q^2}_{(n-1)q^2} + \underbrace{q^2 + \ldots q^2}_{(n-2)q^2} + \ldots + \underbrace{q_{n-1}q_n}_{(n-(n-1))q^2} \\ =&q^2(n-1 + n-2 + \ldots + n - (n-1))\\ =&q^2((n-1)n - 1 - 2 - \ldots - (n-1))\\ =&q^2\left((n-1)n - \sum_{i=1}^{n-1}{i}\right)\\ =&q^2\left((n-1)n - \frac{(n-1)((n-1) + 1)}{2}\right)\\ =&q^2\left((n-1)n - \frac{(n-1)n}{2}\right)\\ =&q^2\frac{(n-1)n}{2} \end{align}

Answer 1

Your approach is fine, also with respect to the usage of an empty sum $\sum_{i=1}^{0}f(i)=0 $.

Using the sigma notation and setting $q:=q_1$ we can write \begin{align*} \sum_{i=1}^n\sum_{j=1}^{i-1}q_iq_j&=\sum_{i=2}^n\sum_{j=1}^{i-1}q_iq_j\tag{1}\\ &=\sum_{i=1}^{n-1}\sum_{j=1}^{i}q_{i+1}q_j\tag{2}\\ &=q^2\sum_{i=1}^{n-1}\sum_{j=1}^{i}1\tag{3}\\ &=q^2\sum_{i=1}^{n-1}i\tag{4}\\ &=q^2\frac{(n-1)n}{2}\tag{5}\\ \end{align*}

Comment:

• In (1) we start with index $i=2$, since the inner sum is empty when $i=1$.

• In (2) we shift the index $i$ to start from $i=1$.

• In (3) we factor out $q$ since $q=q_i=q_j$ for all $i\ne j$.

• In (4) we simplify the inner sum.

• In (5) we simplify the outer sum by applying the corresponding summation formula.