series question

**margaritas** · May 21st 2007, 08:02 PM

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.

**CaptainBlack** · May 21st 2007, 09:45 PM

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.

$\sum_{r=1}^{ n} 3^{r-1}+2r = \left( \sum_{r=1}^{ n} 3^{r-1} \right)+ \left( \sum_{r=1}^{ n} 2r \right)$

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

**margaritas** · May 22nd 2007, 06:53 AM

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

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

**topsquark** · May 22nd 2007, 07:31 AM

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:
$\sum_{k = 0}^n ar^k = \frac{a(1 - r^{n+1})}{1 - r}$

and

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

and you can take it from there.

-Dan

**CaptainBlack** · May 22nd 2007, 07:34 AM

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!

$<br /> \sum_{r=1}^{ n} 3^{r-1} = \frac{1-3^r}{1-3}<br />$

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

RonL

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