Change of summation order in $\sum_{i=0}^n \sum_{j=i}^n a_{j-i}$

Question

I have to show that $$\sum_{i=0}^n \sum_{j=i}^n a_{j-i}$$ is equal to $$\sum_{j=0}^n (n+1-j)a_{j}$$ I can change order of summation or add i to internal sum: $$\sum_{i=0}^n \sum_{j=i}^n a_{j-i} = \sum_{j=0}^n \sum_{i=0}^j a_{j-i} = \sum_{i=0}^n \sum_{j=0}^{n-i} a_{j}$$ But I don't know if it leads somewhere?

Or if I will rewrite this as: $$a_{0} + a_{1} + ... + a_{n}$$ $$+ a_{0} + a_{1} + ... + a_{n-1}$$ $$...$$ $$+ a_{0} + a_{1}$$ $$+ a_{0}$$

And if we'll add columns instead of rows, then sum is $na_{0} + (n-1)a_{1} ... $ etc. But this takes time and I'm looking for faster way, maybe just operating on sigma simbol can lead to this result?

Answer 1

Add $i$ to the internal sum AND change the order of summation:

$$\sum_{i=0}^n \sum_{j=0}^{n-i} a_j = \sum_{j=0}^n \sum_{i=0}^{n-j} a_j = \sum_{j=0}^n (n+1-j)a_j$$

Answer 2

The sum is just $\sum n_j a_j$ where $n_j$ is the number of ways of writing $j$ as $l-m$ with $l \geq m$. [All variable between $0$ and $n$]. There are $n+1-j$ ways of choosing $l \geq j$ and for each such $l$ we have the unique value $m=l-j$.