Proof of a double summation reduction identity 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.
 A: $$\sum_{i=1}^n\sum_{j=1}^nx_{ij}=\sum_{i=1}^n\sum_{j=1}^nx_{ji}\;;\tag{1}$$
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.
A: Hints:
$$\sum_{i=1}^n\sum_{j=1}^ia_j=\sum_{i=1}^n\left(a_1+a_2+\ldots+a_i\right)=a_1+(a_1+a_2)+\ldots+(a_1+\ldots+a_n)=$$
$$na_1+(n-1)a_2+\ldots +(n-(k-1))a_k+\ldots+a_n\;,\;\text{(observe that}\;n-(n-1)=1)$$
A: You can reverse the order of summation in much the same way that one reverses the order of integration in an iterated integral:
$$\sum_{i=1}^n\sum_{j=1}^ia_j=\sum_{j=1}^n\sum_{i=j}^na_j\;;\tag{1}$$
now just notice that the inner summation on the righthand side of $(1)$ is the sum of $(n-j-1)$ copies of $a_j$.
If you've not seen other examples of reversing the order of summation, the array below may be helpful:
$$\begin{array}{c|cccccc|c}
i\backslash j&1&2&3&\dots&n-1&n&\sum_{j=1}^ia_j\\ \hline
1&a_1&&&&&&\sum_{j=1}^1a_j\\
2&a_1&a_2&&&&&\sum_{j=1}^2a_j\\
3&a_1&a_2&a_3&&&&\sum_{j=1}^3a_j\\
\vdots&\vdots&\vdots&\vdots&\ddots&&&\vdots\\
n-1&a_1&a_2&a_3&\dots&a_{n-1}&&\sum_{j=1}^{n-1}a_j\\
n&a_1&a_2&a_3&\dots&a_{n-1}&a_n&\sum_{j=1}^ma_j\\ \hline
\sum_{i=j}^na_j&\sum_{i=1}^na_1&\sum_{i=2}^na_2&\sum_{i=3}^na_3&\dots&\sum_{i=n-1}^na_{n-1}&\sum_{i=n}^na_n&\text{double sum}\atop(1)
\end{array}$$
A: Hint: Try to figure out how many times 


*

*$a_1$ occurs in LHS

*$a_n$ occurs in LHS

*$a_k$, $k \leq n$ occurs in LHS


Guess this will help you out.
