I'll do problem 13 to get you started.
The expression [tex]4\left(\frac{2^n}{7^n}\right)[/tex] is the same as [tex]4\left(\frac{2}{7}\right)^n[/tex]
Then we can do a bit of algebra like so to change that n into n-1
[tex]4\left(\frac{2}{7}\right)^n\\\\4\left(\frac{2}{7}\right)^n*1\\\\4\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{0}\\\\4\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{1-1}\\\\4\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{1}*\left(\frac{2}{7}\right)^{-1}\\\\4*\left(\frac{2}{7}\right)^{1}\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{-1}\\\\\frac{8}{7}\left(\frac{2}{7}\right)^{n-1}\\\\[/tex]
This is so we can get the expression in a(r)^(n-1) form
- a = 8/7 is the first term of the geometric sequence
- r = 2/7 is the common ratio
Note that -1 < 2/7 < 1, which satisfies the condition that -1 < r < 1. This means the infinite sum converges to some single finite value (rather than diverge to positive or negative infinity).
We'll plug those a and r values into the infinite geometric sum formula below
S = a/(1-r)
S = (8/7)/(1 - 2/7)
S = (8/7)/(5/7)
S = (8/7)*(7/5)
S = 8/5
S = 1.6
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Answer in fraction form = 8/5
Answer in decimal form = 1.6
