# Trying to convert a summation to an equation

Here is what I am trying to figure out $\sum_{i=1}^n 3+2 (1-i)$ It would be nice if this could be put into a single equation to be used in a larger system. I know that it is possible to break up the summation into two parts like this $\sum_{i=1}^n 3+\sum_{i=1}^n2 (1-i)$ and it then becomes $3*n+\sum_{i=1}^n2 (1-i)$ I get stuck in trying to convert the second summation into a regular equation. Ultimately, it seems that this should be some sort of exponential function, but I am not having any luck finding it.

This is for a personal project and not homework in case this is a concern.

Brandon

Well, $\displaystyle \sum_{i=1}^n 2(1-i) = \sum_{i=1}^n 2 - 2\sum_{i=1}^n i = 2n - \frac{2n(n+1)}{2}$.

Then we have $\displaystyle \sum_{i=1}^n 3 +2(1-i) = 3n + 2n - n(n+1) = 4n - n^2$.

For intuition on $\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}$, notice $1 + 2 + 3 + \ldots + (n-2) + (n-1) +n = (n+1) + (n-1 +2) + (n -2 +3) + \ldots + (\frac{n}{2} + (\frac{n}{2}+1)) = (n+1) + (n+1) + (n+1) + \ldots + (n+1)$,

grouping the first and last terms together, then the second and second to last, etc., where there are $\frac{n}{2}$ (for even $n$) terms in the last sum. If $n$ is odd, treat with $\frac{n+1}{2}$ where necessary.

Then summing $n+1$ $\frac{n}{2}$ times clearly gives $\frac{n(n+1)}{2}$.

• Awesome explanation. Thank you! – Brandon Dec 29 '16 at 21:02

Hint :- $\sum_{i=1}^n2(1-i)=\sum_{i=1}^n2-2\sum_{i=1}^ni=2.n-2(\frac{n(n+1)}{2})$.