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Prove the identity:

$$\begin{align*} &\sum_{i=0}^{n}\sum_{j=0}^{i} a_ia_j = \frac{1}{2}\left ( \left ( \sum_{i=0}^{n}a_i \right )^2 + \left ( \sum_{i=0}^{n}a_i^2 \right )\right ) \end{align*}$$

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Best answer
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$$\begin{align*} &\quad \;\;\; S = \sum_{{\color{red}{i=0}}}^{n}\sum_{{\color{blue}{j=0}}}^{i} a_{\color{red}{i}}.a_{\color{blue}{j}} \quad \dots \dots (1) \\ \end{align*}$$

From above figure,

$$\begin{align*} &\Rightarrow S = \sum_{{\color{blue}{j=0}}}^{n}\sum_{{\color{red}{i=j}}}^{n} a_{\color{red}{i}}.a_{\color{blue}{j}} \\ \end{align*}$$

Now, $a_j.a_i = a_i.a_j$

$$\begin{align*} &\Rightarrow S = \sum_{{\color{blue}{j=0}}}^{n}\sum_{{\color{red}{i=j}}}^{n} a_{\color{blue}{j}}.a_{\color{red}{i}} \\ \end{align*}$$

Now with change of varibales,

$$\begin{align*} &\Rightarrow S = \sum_{{\color{red}{i=0}}}^{n}\sum_{{\color{blue}{j=i}}}^{n} a_{\color{red}{i}}.a_{\color{blue}{i}} \quad \dots \dots (2)\\ \end{align*}$$

Adding,

$$\begin{align*} &(1)+(2) \rightarrow \\ &\Rightarrow 2S = \sum_{i=0}^{n}\left [ \sum_{j=0}^{i}a_i.a_j \;\; + \;\; \sum_{j=i}^{n}a_i.a_j \right ] \\ &\Rightarrow 2S = \sum_{i=0}^{n}\left [ \sum_{j=0}^{n}a_i.a_j \;\; + \;\; a_i^2 \right ] \\ &\Rightarrow 2S = \sum_{i=0}^{n}\sum_{j=0}^{n}a_i.a_j \;\; + \;\; \sum_{i=0}^{n}a_i^2 \\ &\Rightarrow 2S = \left ( \sum_{i=0}^{n}a_i \right )\left ( \sum_{j=0}^{n}a_j \right ) \;\; + \;\; \sum_{i=0}^{n}a_i^2 \\ &\Rightarrow S = \frac{1}{2}.\left [ \left ( \sum_{i=0}^{n}a_i \right )^2 \;\; + \;\; \left ( \sum_{i=0}^{n}a_i^2 \right ) \right ] \\ \end{align*}$$


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–1 votes
–1 votes
$\sum_{i=0}^{n}\sum_{j=0}^{i} a_ia_j = $

=$\sum_{i=0}^{n}a_{i}\sum_{j=0}^{i}a_{j}$

=$\sum_{i=0}^{n}a_{i}\frac{i(i+1)}{2}$

$=\sum_{i=0}^{n}i\frac{i(i+1)}{2}$

$=\frac{1}{2}\sum_{i=0}^{n}(i^{3}+i^{2})$

$=1^{3}+2^{3}+3^{3}+.................+n^{3}+1^{2}+2^{2}+.................+n^{2}$

$=\left (\frac{n(n+1)}{2} \right) ^{2}$+$\left (\frac{n(n+1)(2n+1)}{6} \right)$

$ = \frac{1}{2}\left ( \left ( \sum_{i=0}^{n}a_i \right )^2 + \left ( \sum_{i=0}^{n}a_i^2 \right )\right ) $
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