Construct a series: {(sigma) n=7k to 7(k+1) a} = 1/(2^k)

[jmarsh]

Hey i have this question from a worksheet that I dont exactly know what there asking so some direction is all i need. The question is:

1. Construct a series such that for each k {(sigma) n=7k to 7(k+1) a} = 1/(2^k)

I hope someone can understand what im trying to get across if someone could also tell me how to post the sigma symbol, integral symbol, subscripts, superscripts, etc. that would be great as well.

 

[Unco]

If that is:\(\displaystyle \large \mbox{ \sum_{n=7k}^{7(k+1)} a = \frac{1}{2^k}}\)
then the sum is just 8a; solve for a.