What is the formula for finding the sum to n terms in a geometric series?

[lionely]

Homework Statement

Wn = 2 + 3(1/2)^n

Homework Equations

The Attempt at a Solution

I am confused, all I tried so far is writing out the first 5 terms, but all that was helping me to do is basically find the Sum to infinity... so what should I do to find the Sum to n terms? I know the 3(1/2)^n will be a g.p. , and the 2 makes it sort of an ap.

 

[Mark44]

Homework Statement

Wn = 2 + 3(1/2)^n

Homework Equations

The Attempt at a Solution

I am confused, all I tried so far is writing out the first 5 terms, but all that was helping me to do is basically find the Sum to infinity... so what should I do to find the Sum to n terms? I know the 3(1/2)^n will be a g.p. , and the 2 makes it sort of an ap.

Since n is going to be a parameter representing the number of terms, it shouldn't also be an index for the individual terms, so write you sequence as W_k = 2 + 3(1/2)^k.

Now, since you want the sum of the first n terms, your summation looks like this:
$$ \sum_{k = 1}^n (2 + 3(1/2)^k)$$

What properties of summations do you know?

 

[lionely]

I can separate the sums , $$ \sum_{k = 1}^n (2 )$$ and $$ \sum_{k = 1}^n (3(1/2)^k)$$

I'm not sure how to separate the latter, never seen one with an index before, well I don't recall..

 

[scurty]

$$ \sum_{k = 1}^n (3(1/2)^k)$$

I'm not sure how to separate the latter, never seen one with an index before, well I don't recall..

Try writing a few terms out and notice what you can do with all the 3s that appear. Do you recall the Geometric Series?

 

[lionely]

I could sum the 3s separately?,Oh and then just replace the SUm of the (1/2)^k with the basic geometric sum formula?

so the Sn = 2n + 3n + a(1-(1/2)^n/[1-1/2] ?

 

[lionely]

nevermind I got it to be 2N + 3(1-2^-N)