Ok, here's the sum.
(7^4) - (7^5) + (7^6) - ... + ((-1)^k) * (7^k) where k is in the set of intergers and k>=4.
All the examples in my book are too different to be of much help to me and I've never done geometric sequence before. Thanks in advance!
Anonymous
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You are given that the general term is (-1)^k * 7^k = (-7)^k (this equality follows from the law of exponents). Assuming that the sum goes from 4 to k inclusively, we can write:
7^4 - 7^5 + 7^6 - ... + (-1)^k * 7^k = ∑ (-7)^n (from n=4 to k).
Now, ∑ (-7)^n (from n=4 to infinity) is a geometric series since the ratio for consecutive terms, (a_n+1)/(a_n), is a constant (-7). The first term of the geometric series is (-7)^4 = 2401 (since the first term occurs at k = 4) and the common ratio is -7. Using the fact that the sum of the first n terms of a geometric series with a first term of a and a common ratio of r is:
a(1 - r^n)/(1 - r),
the required sum is:
∑ (-7)^n (from n=4 to k) = 2401[1 - (-7)^(k - 3)]/[1 - (-7)]
(Note that the series runs from 4 to k, so there are (k - 4) + 1 = k - 3 terms.)
= 2401[1 - (-7)^(k - 3)]/8.
I hope this helps!