How to “make up a formula” of a sum: $S=2+7+12+\cdots+(5n-3)$?

Determine a formula for the following sum with $n\in\mathbb N$:

$$S=2+7+12+\cdots+(5n-3)$$

I had no clue about what to do about it. I just wrote, for the sake of doing something, the following:

$$\sum_{i=0}^n(5i-3) = \left(\sum_{i=0}^n(5i) - 3n\right)$$

Curiously, I got a few points for that. But eventually, there is an annotation by the professor:

And?

So apparently I was on the right path. What was I supposed to do afterward? How do you "make up a formula" of a given sum?

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The sum should be taken from $i = 1$ to $n$, by the way =) – amWhy Dec 1 '12 at 20:44
May I suggest a little reading on this topic? It is the following "tutorial" by D. Gleich: cs.purdue.edu/homes/dgleich/publications/… – Giuseppe Negro Dec 1 '12 at 21:05

So you have $$S = 2 + 7 + 12 + \dots + (5n - 3) = \sum_{i=1}^n (5i-3) = \left(5\sum_{i=1}^{n} i\right) - 3n.$$ So all you need to know is that $$\sum_{i=1}^{n} i = \frac{n(n-1)}{2}.$$

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Hint: you are almost done, you need to take the $5$ out of the sum (its something like $5x+5y=5(x+y)$) and you should recall how to sum $1+\cdots+n$

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I suspect your professor wanted a closed form for this sum (that is, a formula in terms of $n$ only). You can further simplify your formula to $$5\left(\sum_{i=1}^n i \right) - 3n = 5 \frac{n(n-1)}{2} - 3n = \frac{5}{2}n^2 - \frac{11}{2}n.$$ (Also note your summation should start at $i = 1$, not $i = 0$.)

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$$S_n=2 + 7 + 12 + \dots + (5n - 3)$$ is sum of first n terms of arithmetic progression with first term $a_1=2$ and difference $d=5$. From formula $$S_n=\frac{n}{2}(2a_1+(n-1)d)=\frac{n}{2}(4+5n-5)=\frac{n}{2}(5n-1)$$

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