This is a geometric series so we can use the formula for Sn. We have to find a, r, and n. First, notice that with clever factoring we can rewrite the series to look like this: 5(0.2)3 + 5(0.2)3(0.2) + ... + 5(0.2)3(0.2)6. This looks more like the form a + ar1 + ... + arn that we're used to. Now we can see that - r is the ratio between terms, which is 0.2.
- a is the first term, which is 5(0.2)3. This term happens to include a factor of (0.2)3, but that doesn't matter.
- n is the number of terms. There are two ways to count the terms. If we look at the original series,
5(0.2)3 + 5(0.2)4 + ... + 5(0.2)9, we have terms with exponents from 3 to 9, which means we have all the exponents from 1 to 9 except for the first two. This means there are 9 – 2 = 7 terms, so n = 7. If instead we look at the rewritten series, we can see that the exponents on the ratio go from 0 to 6: 5(0.2)3(0.2)0 + 5(0.2)3(0.2)1 + ... + 5(0.2)3(0.2)6 This means we have 7 terms, so n = 7. Finding a, r, and n is most of the work. Once we have those values we stick them in the formula, and stick the formula in the calculator. 
If we're given a sum in sigma notation, it can be helpful to expand the sum first before we go hunting for r, a, and n. | 
