geometric series
- Thread starter thebenji
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\(\displaystyle \L\\\sum_{n=0}^{\infty}\frac{2^{n}}{4^{2n+1}}\)
\(\displaystyle \L\\\frac{2^{n}}{4^{2n+1}}=\frac{2^{n}}{4^{2n}\cdot{4}}=\frac{2^{n}}{16^{n}\cdot{4}}=\frac{1}{4}\cdot\frac{1}{8^{n}}\)
Factor out the 1/4:
\(\displaystyle \L\\\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{8^{n}}\)
Geometric series: \(\displaystyle \L\\\frac{1}{1-\frac{1}{8}}=\frac{8}{7}\)
\(\displaystyle \L\\\frac{8}{7}\cdot\frac{1}{4}=\frac{2}{7}\)
\(\displaystyle \L\\\frac{2^{n}}{4^{2n+1}}=\frac{2^{n}}{4^{2n}\cdot{4}}=\frac{2^{n}}{16^{n}\cdot{4}}=\frac{1}{4}\cdot\frac{1}{8^{n}}\)
Factor out the 1/4:
\(\displaystyle \L\\\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{8^{n}}\)
Geometric series: \(\displaystyle \L\\\frac{1}{1-\frac{1}{8}}=\frac{8}{7}\)
\(\displaystyle \L\\\frac{8}{7}\cdot\frac{1}{4}=\frac{2}{7}\)
skeeter
Elite Member
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- Dec 15, 2005
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start by "rearranging" the denominator ...
4<sup>2n+1</sup> = 2<sup>4n+2</sup> = 2<sup>4n</sup>*2<sup>2</sup> = 4*(2<sup>4</sup>)<sup>n</sup> = 4*16<sup>n</sup>
2<sup>n</sup>/(4*16<sup>n</sup>) = (1/4)(2<sup>n</sup>/16<sup>n</sup>) = (1/4)(1/8)<sup>n</sup>
now you have an infinite geometric series whose sum is of the form ... k*a<sub>o</sub>/(1 - r)
S = (1/4)*1/(1 - 1/8) = (1/4)(8/7) = 2/7
4<sup>2n+1</sup> = 2<sup>4n+2</sup> = 2<sup>4n</sup>*2<sup>2</sup> = 4*(2<sup>4</sup>)<sup>n</sup> = 4*16<sup>n</sup>
2<sup>n</sup>/(4*16<sup>n</sup>) = (1/4)(2<sup>n</sup>/16<sup>n</sup>) = (1/4)(1/8)<sup>n</sup>
now you have an infinite geometric series whose sum is of the form ... k*a<sub>o</sub>/(1 - r)
S = (1/4)*1/(1 - 1/8) = (1/4)(8/7) = 2/7