summation
I'd like to see a proof of this identity I discovered:
$$
\sum_{i=1}^{n}\sum_{j=1}^{i}a_j = \sum_{i=1}^{n}a_i(n-i+1)
$$
Probably there's something I forgot because I couldn't manage it.
[edit]Possibly with just identities, not examples please!
[edit2]No induction procedure, either.
\begin{align} \left( \sum_{i=1}^n|a_i|^2 \right)\left( \sum_{i=1}^n |b_i|^2 \right)-\left| \sum_{i=1}^na_ib_i \right|^2&=\sum_{i,j=1}^n|a_i|^2|b_j|^2 -\sum_{i,j=1}^n\bar{a}_i\bar{b}_ia_jb_j \\ &=\sum_{i,j=1, i\neq j}^n|a_i|^2|b_j|^2 +\sum_{i=1}^n|a_i|^2|b_i|^2 -\sum_{i=1}^n|a_i|^2|b_i|^2 \\ & \hspace{5 mm}-\sum_{i,j=1, i\neq j}^n\bar{a}_i\bar{b}_ia_jb_j \\ &=\sum_{i,j=1, i\neq j}^n|a_i|^2|b_j|^2 -\sum_{i,j=1, i\neq j}^n\bar{a}_i\bar{b}_ia_jb_j \\ &=\dfrac{1}{2}\left(\sum_{i,j=1, i\neq j}^n |a_i|^2|b_j|^2+\sum_{i,j=1, i\neq j}^n |a_j|^2|b_i|^2\right) - \\&\quad\quad \dfrac{1}{2}\left(\sum_{i,j=1, i\neq j}^n \bar{a}_i\bar{b}_ia_jb_j+\sum_{i,j=1, i\neq j}^n \bar{a}_j\bar{b}_ja_ib_i\right)\:\text{*} \\ &=\dfrac{1}{2}\sum_{i,j=1, i\neq j}^n(|a_i|^2|b_j|^2+|a_j|^2|b_i|^2 -\bar{a}_i\bar{b}_ia_jb_j-\bar{a}_j\bar{b}_ja_ib_i) \\ &=\dfrac{1}{2}\sum_{i,j=1, i\neq j}^n(\bar{a}_ib_j-\bar{a}_jb_i)(a_i\bar{b}_j-a_j\bar{b}_i) \\ &=\dfrac{1}{2}\sum_{i,j=1, i\neq j}^n|\bar{a}_ib_j-\bar{a}_jb_i|^2 \\ &=\dfrac{1}{2}\left(\sum_{1\leqslant i<j\leqslant n}|\bar{a}_ib_j-\bar{a}_jb_i|^2+\sum_{1\leqslant j<i\leqslant n}|\bar{a}_ib_j-\bar{a}_jb_i|^2\right) \\ &=\dfrac{1}{2}\left(\sum_{1\leqslant i<j\leqslant n}|\bar{a}_ib_j-\bar{a}_jb_i|^2+\sum_{1\leqslant i<j\leqslant n}|\bar{a}_jb_i-\bar{a}_ib_j|^2\right) \\ &=\sum_{1\leqslant i<j\leqslant n}|\bar{a}_ib_j-\bar{a}_jb_i|^2 \\ \end{align} * Since $$ \quad \sum \limits_{i,j=1, i\neq j}^n\bar{a}_i\bar{b}_i a_j b_j=\left| \sum_{i=1}^na_ib_i \right|^2-\sum_{i=1}^n|a_i|^2|b_i|^2 $$ It is real. So $$ \overline{\sum \limits_{i,j=1, i\neq j}^n\bar{a}_i\bar{b}_ia_jb_j}=\sum \limits_{i,j=1, i\neq j}^n a_ib_i\bar{a}_j\bar{b}_j=\sum \limits_{i,j=1, i\neq j}^n\bar{a}_i\bar{b}_ia_jb_j $$
The best way to understand it is to draw the set of couples $(i,j)$ such that
$$k\le j\le i\le n$$ as this
The cases where there are the stars are the set of desired couples and for example the first equality is to sweep these couples by rows: $$\underbrace{(a_{k,k})}_{\text{first row}}+(a_{k+1,k}+a_{k+1,k+1})+\cdots+\underbrace{(a_{n,k}+a_{n,k+1}+\cdots+a_{n,n})}_{\text{last rows}}$$
Best Answer
$$\sum_{i=1}^n\sum_{j=1}^nx_{ij}=\sum_{i=1}^n\sum_{j=1}^nx_{ji}\;;\tag{1}$$ because each summand $x_{rs}$ on the RHS can be found on the LHS and vice versa and summation is commutativ. Therefore this sum is often written as
$$ \sum_{1\le i,j\le n}x_{ij}$$
We set
$$ x_{rs}=a_s,s \le r \;;\tag{2}$$ $$ x_{rs}=0, s \gt r \;;\tag{3}$$
Expanding the LHS from (1) and substituting (2) and (3) we get
$$\sum_{i=1}^n\sum_{j=1}^nx_{ij}=\sum_{i=1}^n\sum_{j=1}^{i}x_{ij}+\sum_{i=1}^n\sum_{j=i+1}^nx_{ij}=\sum_{i=1}^n\sum_{j=1}^{i}a_j+\sum_{i=1}^n\sum_{j=i+1}^n0=\sum_{i=1}^n\sum_{j=1}^ia_j$$
This is the LHS of your idendity.
Expanding the RHS of (1) and substituting (2) and (3) gives
$$\sum_{i=1}^n\sum_{j=1}^nx_{ji}=\sum_{i=1}^n\sum_{j=1}^{i-1}x_{ji}+\sum_{i=1}^n\sum_{j=i}^nx_{ji}=\sum_{i=1}^n\sum_{j=1}^{i-1}0+\sum_{i=1}^n\sum_{j=i}^na_i=\sum_{i=1}^na_i(\sum_{j=i}^n1)=\sum_{i=1}^na_i(\sum_{j=1}^n1-\sum_{j=1}^{i-1}1)=\sum_{i=1}^na_i(n-(i-1))$$
This is the RHS of your idendity.