Htykuut

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 $deltainmathbb{N}$:

$$sum_{n=1}^{50}2n+delta$$

where $delta$ is the current sum for each $n$.
So to avoid confusion the actual computation that I want is the following:

$$
begin{split}
2cdot1+0&=2\
2cdot2+2&=6\
2cdot3+6&=12\
2cdot4+12&=20\
2cdot5+20&=30\
2cdot6+30&=42\
2cdot7+42&=56\
2cdot8+56&=72\
2cdot9+72&=90\text{etc..}
end{split}
$$

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, $a(n) = ncdot(n+1).$ 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
$$a_k = displaystylesum_{n=1}^{k}2n+a_{n-1}$$
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
$$
begin{split}
a_n &= sum_{k=1}^n 2k + a_0 \
&= 2sum_{k=1}^n k \
&= 2 cdot frac{n(n+1)}{2} \
&= n(n+1).
end{split}
$$

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

• if $n$ is even, there are exactly $n/2$ such pairs, so the sum is $(n+1)n/2$
  


• if $n$ is odd, there are $(n-1)/2$ such pairs and the middle number is $(n+1)/2$, so the sum is
  $$
  frac{(n+1)(n-1)}{2} + frac{n+1}{2}
  = frac{n+1}{2} left[(n-1)+1right]
  = frac{n(n+1)}{2}.
  $$
  

Doubly, or triply, (or morely) applied sum symbols are just a normal occurence, & something to get used-to.