I am only confused about the series part of this question. The solution to that part was:
$2V^8+2V^{16}+....... =2 V^8/(1-V^8)$
How did they get this, is it a geometric series rule?
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I did it in 2nd year, but I just was not connecting it to this until now. a=2V^8 and r=V^8 – Mi-lee Wilson Sep 13 '15 at 17:53
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Yes. The sum of the geometric series
$$\displaystyle\sum_{n=0}^{\infty} ar^n$$
for $|r| < 1$ is given by
$$\frac{a}{1-r}$$
In your example, we have $a = 2V^8$ and $r = V^8$
See Geometric series for more information.
Sam Cappleman-Lynes
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@Mi-leeWilson IMO the right way to thank on MSE is to upvote/accept answer. – Arpit Kansal Sep 13 '15 at 17:51