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

## 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 $\dfrac 1 2$.

$\blacksquare$