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leonid [27]
3 years ago
8

Find the SUm of the infinite series.

Mathematics
1 answer:
givi [52]3 years ago
7 0

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

------------------------

Answer in fraction form = 8/5

Answer in decimal form = 1.6

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Answer:

x intercept = -3

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The x-intercepts are where the graph crosses the x-axis

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If the second term of a geometric sequence of real numbers is -2 and the fifth term is 16 then what is the fourteenth term?
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<h3>Given</h3>

A geometric sequence such that ...

a_2=-2\quad\text{and}\quad a_5=16

<h3>Find</h3>

a_{14}

<h3>Solution</h3>

We can use the ratio of the given terms to find the common ratio of the sequence, then use that to find the desired term from one of the given terms. We don't actually need the common ratio (-2). All we need is its cube (-8).

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