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How can we show that this double sum is equal to product of single sum?

$$\sum_{i=0}^{n}\sum_{j=0}^{n}a_ia_j = \left(\sum_{i=0}^{n}a_i\right)^2$$

I tried like this, that for the inner sum, we can take $a_i$ common:

$$\sum_{i=0}^{n}\left[a_i\left(\sum_{j=0}^{n}a_j\right)\right]$$ Then for outer sum we can take $\left(\sum_{j=0}^{n}a_j\right)$ common to get:

$$\left(\sum_{i=0}^{n}a_i\right)\times\left(\sum_{j=0}^{n}a_j\right) = \left(\sum_{i=0}^{n}a_i\right)^2$$

Is this method ok? Also give some insights on double sums and how to evaluate them! Thanks a lot!

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    $\begingroup$ Your reasoning looks perfectly sound to me. $\endgroup$ – TonyK Jun 26 '17 at 11:06
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    $\begingroup$ Your argument works perfectly. Good job. $\endgroup$ – Crostul Jun 26 '17 at 11:07
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    $\begingroup$ Hint: Proof by induction on $n$. $\endgroup$ – TheGeekGreek Jun 26 '17 at 11:08
  • $\begingroup$ Thanks! Out teacher directly told a this property so I couldnt understand. But now I get it! $\endgroup$ – akhmeteni Jun 26 '17 at 11:18
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You might want to think about it combinatorially: when you multiply $$(a_1+\cdots + a_n)(a_1+\cdots + a_n)$$ you actually choose one term from the left hand side, multiply it by another term from the r.h.s, and sum all of it up. Thus you have a sum of terms of the form $a_ia_j$ where $1\leq i,j\leq n$, so you can write: $$(a_1+\cdots + a_n)^2=\sum_i\sum_ja_ia_j$$ Similarly, you can form triple sums by looking at $$(a_1+\cdots + a_n)(a_1+\cdots + a_n)(a_1+\cdots + a_n)$$ and this will be a sum of $a_ia_ja_k$ where $1\leq i,j,k\leq n$, so $$(a_1+\cdots + a_n)^3=\sum_i\sum_j\sum_ka_ia_ja_k,$$ and so on.

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  • $\begingroup$ Yes I tried to write terms and this too verified that! Thanks! $\endgroup$ – akhmeteni Jun 26 '17 at 11:19
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hint

$$\sum_{i=0}^n(\sum_{j=0}^na_ia_j )=$$

$$\sum_{i=0}^n(a_i\sum_{j=0}^na_j) $$

$$a_0+a_1+...a_n=\sum_{k=0}^n a_k $$

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  • $\begingroup$ The OP already said that... $\endgroup$ – TonyK Jun 26 '17 at 11:05
  • $\begingroup$ And it looks like the OP already knows that too... $\endgroup$ – TonyK Jun 26 '17 at 11:07
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$\sum\limits_{i=0}^{n}\sum\limits_{j=0}^{n}a_ia_j = \sum\limits_{i=0}^{n}\left(a_ia_0+\dots +a_ia_n\right)=\left(a_0a_0+\dots +a_0a_n\right)+\dots +\left(a_na_0+\dots +a_na_n\right)=\left(a_0+\dots +a_n\right)^2=\left(\sum\limits_{i=0}^{n}a_i\right)^2$

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