summing geometric series

You've pretty much got it. My advice is to factor in the manner I've shown below and always get the series into the for a ( r^0 + r^1 + ... + r^n). See my notes.

Also send me your access code so I can post this to your page. I'll be sending you a new 7-digit code and copying all your existing stuff to the new location.

I was doing some revision for Test # 4 and was going over Geometric series.

I was playing around with some numbers and have now confused myself…

Example: I was finding the sum of the first (x) number of terms in a series, and in an effort to prove I was doing the right thing I set up the following:

4/9 +16/9 + 64/9 + 256/9 . As you know this is in the form (ar^1+ar^2+ar^3+ar^4)

Now the sum of (4/9 +16/9 + 64/9 + 256/9) = 37.77777

However, when I key in a( (1-r^4)/(1-r) ) I get 9.4444 using r=4 and a=(1/9)

The formula is
a ( r^0 + r^1 + r^2 + ... + r^n ) = ( 1 - r^(n+1) ) / (1 - r).
If you factor out 4/9 you get
4/9 (1 + 4 + 16 + 64) .
If you let a = 4/9 and r = 4, this is of the form
a ( r^0 + r^1 + ... + r^n)
with n = 3, so the sum would be
a ( 1 - r^(n+1) ) / ( 1 - r ) = 4/9 ( 1 - 4^4) / ( 1 - 4) = 4/9 * (-255) / (-3).
I believe this does come out to 37.777.

I thought I understood that the important thing was to identify where the series started and how to apply the Sum equation. Ie , the above starts at (a*r^1) thus solving for (a) and (r ) then applying a( (1-r^n)/(1-r) ) . Had it started at (a*r^0) I would have used a( (1-r^n+1)/(1-r) ).

I hope all is well with you and your Family. I hope to test Friday or Monday, thus completing the course…