Question

Find the SUm of the infinite series.

Answer 1

I'll do problem 13 to get you started.

The expression $4\left(\frac{2^n}{7^n}\right)$ is the same as $4\left(\frac{2}{7}\right)^n$

Then we can do a bit of algebra like so to change that n into n-1

$4\left(\frac{2}{7}\right)^n\\\\4\left(\frac{2}{7}\right)^n*1\\\\4\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{0}\\\\4\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{1-1}\\\\4\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{1}*\left(\frac{2}{7}\right)^{-1}\\\\4*\left(\frac{2}{7}\right)^{1}\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{-1}\\\\\frac{8}{7}\left(\frac{2}{7}\right)^{n-1}$

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