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$$\text{Assertion:}\;\left(\sum_{i=1}^n |a_i|\right)^2\overset{(*)}{=}\sum_{i=1}^n |a_i|^2+2\sum_{i<j}^n |a_i|\cdot|a_j|$$

whereby $i,j\in\mathbb{N}$ and $a\in\mathbb{R}$. First I'll write what I've accomplished so far:

Let $n=2$ (base clause), then $(\sum_{i=1}^2 |a_i|)^2=(|a_1|+|a_2|)^2=|a_1|^2+|a_2|^2+2|a_1|\cdot|a_2|$. (so the base clause is true).

Now assume that for $n = k$, that $(*)$ holds (induction hypothesis).

$$ \begin{align} n\rightarrow n+1: \left(\sum_{i=1}^{n+1} |a_i|\right)^2 &=\left(\sum_{i=1}^{n+1} |a_i|\right)\left(\sum_{i=1}^{n+1} |a_i|\right)\\ &=\left(|a_{n+1}|+\sum_{i=1}^n |a_i|\right)\left(|a_{n+1}|+\sum_{i=1}^n |a_i|\right)\\ &=|a_{n+1}|^2+2|a_{n+1}|\left(\sum_{i=1}^{n} |a_i|\right)+\left(\sum_{i=1}^{n} |a_i|\right)^2\\ &\overset{(*)}{=}|a_{n+1}|^2+2|a_{n+1}|\left(\sum_{i=1}^{n} |a_i|\right)+\sum_{i=1}^n |a_i|^2+2\sum_{i<j}^n |a_i|\cdot|a_j|\\ &=\sum_{i=1}^{n+1} |a_i|^2+2\left(\sum_{i=1}^n |a_i|\cdot|a_{n+1}|+\sum_{i<j}^n |a_i|\cdot|a_j|\right)\\ \end{align} $$

and that's where I got stuck. Any ideas how to proceed or did I something wrong/need to reconsider?

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You did all the dirty work correctly. Now just notice that the sums in the brackets can be combined into the final form of the sum that you need.

In other words,

$$ \sum_{i=1}^n |a_i|\cdot|a_{n+1}|+\sum_{i<j}^n |a_i|\cdot|a_j| = \sum_{i<j}^{n+1} |a_i|\cdot|a_j| $$

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  • $\begingroup$ I also thought that I had to do just that but I couldn't imagine why that would be possible in the first place. $|a_{n+1}|$ is a constant first right? How did you got the $j$ into the index? $\endgroup$ – 冬海愛衣 Dec 13 '16 at 16:54
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    $\begingroup$ @Stefan $$ \sum_{i=1}^n |a_i|\cdot|a_{n+1}| = |a_{n+1}|\sum_{j=1}^n |a_j| $$ $\endgroup$ – gt6989b Dec 13 '16 at 16:55

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