Question #d27c8

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Question #d27c8

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Answer 1 · 2016-03-21T00:20:49

a. $\infty \sum n = 0 {0.4}^{n} = \frac{5}{3}$ $sum_(n=0)^oo 0.4^n =5/3$

b. $\infty \sum n = 1 \frac{{(- 0.1)}^{n}}{3} = - \frac{1}{33}$ $sum_(n=1)^oo(-0.1)^n/3=-1/33$

Explanation:

A geometric series is a series of the form $\sum n a r^{n}$ $sum_n ar^n$ . If $| r | < 1$ $|r| < 1$ , then we have $\infty \sum n = 0 a r^{n} = \frac{a}{1 - r}$ $sum_(n=0)^ooar^n = a/(1-r)$
Both of the series in question are geometric series.

a. $\infty \sum n = 0 {0.4}^{n} = \frac{1}{1 - 0.4} = \frac{5}{3}$ $sum_(n=0)^oo 0.4^n = 1/(1-0.4) = 5/3$

b. $\infty \sum n = 1 \frac{{(- 0.1)}^{n}}{3} = - \frac{1}{3} + \infty \sum n = 0 \frac{{(- 0.1)}^{n}}{3}$ $sum_(n=1)^oo(-0.1)^n/3 = -1/3 + sum_(n=0)^oo(-0.1)^n/3$

$= - \frac{1}{3} + \frac{1}{3} \infty \sum n = 0 {(- 0.1)}^{n}$ $=-1/3+1/3sum_(n=0)^oo(-0.1)^n$

$= - \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{1 - (- 0.1)}$ $=-1/3 + 1/3*1/(1-(-0.1))$

$= - \frac{1}{3} + \frac{1}{3} \cdot \frac{10}{11}$ $=-1/3 + 1/3*10/11$

$= - \frac{1}{33}$ $=-1/33$

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