OpenStudy (anonymous):

I don't understand this problem. find ∑[i=1,∞,6(3)^(i-1)]

OpenStudy (anonymous):

\[\sum_{i=1}^\infty{6(3)^(i-1)}\]

OpenStudy (anonymous):

is that to the power of i-1?

OpenStudy (anonymous):

hello?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

K so you can observe that this is a geometric series and can be simplified to look like this:\[\bf \sum_{i=1}^{\infty}6(3)^{i-1}=\sum_{i=1}^{\infty}6\frac{(3)^{i}}{3}=\sum_{i=1}^{\infty}2(3)^{i}\]It should be noticed that this is a geometric series with a comman ratio of r = 3. For this series to be convergent, i.e. for it to have a definable sum, the common ratio must be in the interval |r| < 1. Since this isn't the case, the series is divergent and hence the sum cannot be determined. @racsal36

OpenStudy (mertsj):

@genius12 The asker is not too interested, I would say. Apparently assigned his/her work to you and went out to enjoy the Memorial Day weekend. So snap to and get it done!!!