Thread: Infinite Series

I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it.


infinity
∑ (2^(j+4)) / (7^(j-3))
j=6


is there a certain formula i need to follow?
Thanks.

Originally Posted by stevo25

I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it.


infinity
∑ (2^(j+4)) / (7^(j-3))
j=6


is there a certain formula i need to follow?
Thanks.

. Look familiar?

Originally Posted by stevo25

I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it.


infinity
∑ (2^(j+4)) / (7^(j-3))
j=6


is there a certain formula i need to follow?
Thanks.

Now we're left with a geometric series. Can you take it from here?

EDIT: Ah! Drexel28 beat me to the chase!

Ah, it is pretty simple! After that you just add 6 to the j to set the index to 0 and then put (2/7)^6 outside the summation. Then you can plug it into

(ar^(n+1) - a) / (r-1)

formula and take the limit as it approaches infinity.

Thanks!