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.

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- Mar 29th 2010, 06:49 PMstevo25Infinite 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. - Mar 29th 2010, 07:04 PMDrexel28
- Mar 29th 2010, 07:08 PMchiph588@
$\displaystyle \sum_{j=6}^\infty \frac{2^{j+4}}{7^{j-3}} = 2^4\cdot7^3\sum_{j=6}^\infty \frac{2^{j}}{7^{j}} = 2^4\cdot7^3\sum_{j=6}^\infty\left(\frac27\right)^j $

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

**EDIT:**Ah!**Drexel28**beat me to the chase! - Mar 29th 2010, 08:10 PMstevo25
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!