Manipulation of sum

Question

Prove the identity:

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

Answer 1

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

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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Comments

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

how it will be equal to $\sum_{i=0}^{n}\sum_{j=0}^{n}a_{i}a_{j}$

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$2S = \sum_{i=0}^{n}\left [ \sum_{j=0}^{n}a_i.a_j ... \right ]$

plz say how limit changing here?

Answer 2

$\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 ) $

Comments

:) you deliberately assumed $a_i$ as $i$ and replaced $i$ as $a_i$ again. Method is not correct :)

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Can u elaborate more? $a_i$ can be assumed as i
I am not getting u

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We cannot replace $a_i$ by $i$.

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if I replace $i$ with $a_{i}$ then it will be correct.

rt?

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Srestha, you proved a case when $a_i = i$. But $a_i$ can be anything like $a_i = 3 i^2 - 2i$. So, you have to prove it in general.