Summation using previous sum inside the sigma

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)

Is the following notation $$a_k = \displaystyle\sum_{n=1}^{k}2n+a_{n-1}$$ a valid one?

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

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 $$\frac{(n+1)(n-1)}{2} + \frac{n+1}{2} = \frac{n+1}{2} \left[(n-1)+1\right] = \frac{n(n+1)}{2}.$$
• 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? – Natural Number Guy Nov 21 '18 at 23:12
• oops. should be $a_n$ and $a_{n-1}$ instead of $i$. – Natural Number Guy Nov 21 '18 at 23:20
• @NaturalNumberGuy see update – gt6989b Nov 22 '18 at 5:05

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