Summation 6, (1/3)^i+1, i=1

[trip20]

Last question, promise!

Not sure how to write out a summation but here it goes:


6
E (1/3)^i+1
i=1

I know this means: (1/3)^(6+1)+(1/3)^(5+1)+(1/3)^(4+1) .....(1/3)^(1+1)
but I'm not sure how to write this out and solve?

 

[Unco]

Again with the parentheses... anyway, look up the sum of a geometric series.