This is a geometric series so we can use the formula for *S*_{n}. 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* + *ar*^{1} + ... + *ar*^{n}
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*. |