Double summation problem $\sum_{i=1}^{n-1} \sum_{j=i+1}^n X_i X_j.$

Question

I have to calculate this double summation but I am not sure I am doing it the correct way. Could you please help me with it?

The equation is: $$\sum_{i=1}^{n-1} \sum_{j=i+1}^n X_i X_j. $$

So, for example, if I have $\boldsymbol{n=4}$, with $$m_1:=X_1; \quad m_2:=X_2; \quad m_3:=X_3; \quad m_4:=X_4;$$ the equation then becomes $$\sum_{i=1}^{3} \sum_{j=i+1}^4 X_i X_j = (m_1*m_2)+(m_2*m_3)+(m_3*m_4)+(m_2*m_1)\\{}+(m_2*m_2)+(m_2*m_3)+(m_3*m_1)+(m_3*m_2)+(m_3*m_3) $$

Please correct me if I am wrong and thanks in advance for your help.

Answer 1

What is $X_0$?

But finish steeping through all the values of $j$ before incrementing $i.$

$m_1m_2 + m_1m_3 + m_1m_4 + m_2m_3 + m_2m_4 + m_3m_4$

Answer 2

$\def\rj{\color{red}}$ $\def\az{\color{blue}}$ $\def\nr{\color{red}}$ $\def\pp{\color{blue}}$ \begin{align} \sum_{\az{i=1}}^{\az{n-1}} \rj{f(\az{i})} &= \rj{f(\az{1})} +\rj{f(\az{2})} +\ldots +\rj{f(\az{n-2})} +\rj{f(\az{n-1})}.\\ \sum_{\az{i=1}}^{\az{3}} \rj{\sum_{j=\az{i}+1}^{4} X_{\az{i}} X_j} &= \left(\rj{\sum_{j=\az{1}+1}^{4} X_{\az{1}} X_j}\right) +\left(\rj{\sum_{j=\az{2}+1}^{4} X_{\az{2}} X_j}\right) +\left(\rj{\sum_{j=\az{3}+1}^{4} X_{\az{3}} X_j}\right)\\[0.5em] &= \left(\sum_{\pp{j=2}}^{\pp{4}} \nr{X_1 X_{\pp{j}}}\right) +\left(\sum_{\pp{j=3}}^{\pp{4}} \nr{X_2 X_{\pp{j}}}\right) +\left(\sum_{\pp{j=4}}^{\pp{4}} \nr{X_3 X_{\pp{j}}}\right)\\ &= (\nr{X_{1} X_{\pp{2}}} +\nr{X_{1} X_{\pp{3}}} +\nr{X_{1} X_{\pp{4}}}) +(\nr{X_{2} X_{\pp{3}}} +\nr{X_{2} X_{\pp{4}}}) +(\nr{X_{3} X_{\pp{4}}}). \end{align}