Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2/Proof 1

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Theorem

$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \dfrac 1 2 \paren {\paren {\sum_{i \mathop = a}^b x_i}^2 + \paren {\sum_{i \mathop = a}^b {x_i}^2} }$


Proof

Let:

\(\ds S_1\) \(:=\) \(\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = a}^b \sum_{i \mathop = j}^b x_i x_j\) Summation of i from 1 to n of Summation of j from 1 to i
\(\ds \) \(=\) \(\ds \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b x_i x_j\) Change of Index Variable of Summation
\(\ds \) \(=:\) \(\ds S_2\)


Then:

\(\ds 2 S_1\) \(=\) \(\ds S_1 + S_2\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = a}^b \paren {\sum_{j \mathop = a}^i x_i x_j + \sum_{j \mathop = i}^b x_i x_j}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = a}^b \paren {\paren {\sum_{j \mathop = a}^b x_i x_j} + x_i x_i}\) Sum of Summations over Overlapping Domains: Example
\(\ds \) \(=\) \(\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^b x_i x_j + \sum_{i \mathop = a}^b x_i x_i\) Sum of Summations equals Summation of Sum
\(\ds \) \(=\) \(\ds \paren {\sum_{i \mathop = a}^b x_i} \paren {\sum_{j \mathop = a}^b x_j} + \paren {\sum_{i \mathop = a}^b {x_i}^2}\) General Distributivity Theorem and Change of Index Variable of Summation

The result follows on multiplying by $\dfrac 1 2$.

$\blacksquare$


Sources

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