I'm doing a summation, but I need the current sum to be a part of the computation in the actual sigma. First I define $n$ and $\delta\in\mathbb{N}$:
$$\sum_{n=1}^{50}2n+\delta$$
where $\delta$ is the current sum for each $n$. So to avoid confusion the actual computation that I want is the following:
$$ \begin{split} 2\cdot1+0&=2\\ 2\cdot2+2&=6\\ 2\cdot3+6&=12\\ 2\cdot4+12&=20\\ 2\cdot5+20&=30\\ 2\cdot6+30&=42\\ 2\cdot7+42&=56\\ 2\cdot8+56&=72\\ 2\cdot9+72&=90\\\text{etc..} \end{split} $$
You see the previous summation are the $\delta$ in the next summation. As a side note: After just plugging in these numbers in OEIS, I found out that they are the pronic numbers, $a(n) = n\cdot(n+1).$ However, I could have used a totally different example. My question asks wether there is a notation for the sigma summation to get $\delta$ regardless of the outcome?
One idea I have is to use two sigmas instead of one, but I do not want to overcomplicate things if there is a better way..
(Im not necessarily interested in the result of the above computation, only in the process of how the notation works on how to compute)
Update: After some answers and comments, my question can be addressed more clearly:
Is the following notation $$a_k = \displaystyle\sum_{n=1}^{k}2n+a_{n-1}$$ a valid one?
HINT
What you actually are defining is the sequence $a_n$ which satisfies the initial condition $a_0=0$ and recurrence relation $a_n = a_{n-1} + 2n$. Can you now solve it?
UPDATE
I am saying that if you define your function as I am suggesting, $a_0 = \delta$ and $$ \begin{split} a_n &= \sum_{k=1}^n 2k + a_0 \\ &= 2\sum_{k=1}^n k \\ &= 2 \cdot \frac{n(n+1)}{2} \\ &= n(n+1). \end{split} $$
That the formula $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ holds can be proven by noticing that the sum $$ 1 + 2 + 3 + \ldots + (n-2) +(n-1)+n $$ can be grouped into pairs, adding first and last elements together, then second and next-to-last, etc. Each such pair has a sum of $n+1$ and