1
$\begingroup$

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.

Thank you in advance for your help.

Brandon

$\endgroup$
1
$\begingroup$

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 $\frac{n(n+1)}{2}$.

$\endgroup$
  • $\begingroup$ Awesome explanation. Thank you! $\endgroup$ – Brandon Dec 29 '16 at 21:02
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.