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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?

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    $\begingroup$ es, that notation is perfectly valid - that's exactly how I've seen similar things written. $\endgroup$ – Deusovi Nov 26 '18 at 19:49
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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

  • 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 $$ \frac{(n+1)(n-1)}{2} + \frac{n+1}{2} = \frac{n+1}{2} \left[(n-1)+1\right] = \frac{n(n+1)}{2}. $$
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  • $\begingroup$ No. Im not used to mathematical notation. Are you saying that $$\sum_{n=1}^{50}a_i=2n+a_{i-1}$$ is allowed? And also is $a_{i-1}$ always initially 0? $\endgroup$ – Natural Number Guy Nov 21 '18 at 23:12
  • $\begingroup$ oops. should be $a_n$ and $a_{n-1}$ instead of $i$. $\endgroup$ – Natural Number Guy Nov 21 '18 at 23:20
  • $\begingroup$ @NaturalNumberGuy see update $\endgroup$ – gt6989b Nov 22 '18 at 5:05
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Doubly, or triply, (or morely) applied sum symbols are just a normal occurence, & something to get used-to.

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