# Thread: series question

1. ## series question

Hey guys I can't seem to get the answer for this so thanks if anybody could help:

The rth term of a series is 3^(r-1)+2r. Find the sum of the first n terms.

2. Originally Posted by margaritas
Hey guys I can't seem to get the answer for this so thanks if anybody could help:

The rth term of a series is 3^(r-1)+2r. Find the sum of the first n terms.
$\displaystyle \sum_{r=1}^{ n} 3^{r-1}+2r = \left( \sum_{r=1}^{ n} 3^{r-1} \right)+ \left( \sum_{r=1}^{ n} 2r \right)$

The first of these is a finite geometric series which you should be able to sum
and the second is a finite arithmetic series which you should also be able to sum

RonL

3. No, actually I still don't get it. Could you elaborate further? Thanks!

EDIT: Oh no wait, I think I get it!

4. Originally Posted by margaritas
No, actually I still don't get it. Could you elaborate further? Thanks!

EDIT: Oh no wait, I think I get it!
For the record:
$\displaystyle \sum_{k = 0}^n ar^k = \frac{a(1 - r^{n+1})}{1 - r}$

and

$\displaystyle \sum_{k = 0}^n k = \frac{n(n+1)}{2}$

and you can take it from there.

-Dan

5. Originally Posted by margaritas
No, actually I still don't get it. Could you elaborate further? Thanks!

EDIT: Oh no wait, I think I get it!
$\displaystyle \sum_{r=1}^{ n} 3^{r-1} = \frac{1-3^r}{1-3}$

$\displaystyle \sum_{r=1}^{ n} 2r = r(r+1)$

RonL