# Proof by induction of the sum $(a+b+c+d+…)^2$

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

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?

$$\sum_{i=1}^n |a_i|\cdot|a_{n+1}|+\sum_{i<j}^n |a_i|\cdot|a_j| = \sum_{i<j}^{n+1} |a_i|\cdot|a_j|$$
• 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? – 冬海愛衣 Dec 13 '16 at 16:54
• @Stefan $$\sum_{i=1}^n |a_i|\cdot|a_{n+1}| = |a_{n+1}|\sum_{j=1}^n |a_j|$$ – gt6989b Dec 13 '16 at 16:55