I'm struggling with the following sum:
I've tried using a geometric series with the nth term equalingwhere
, and using the geometric sum of the first N+2 terms:
but I don't think thats correct. I just checked in Wolfram Alpha and the answer there is mine multiplied byso I'm probably just missing something simple.
Not really sure what to do really so any help would be appreciated.
I don't think you have applied the formula correctly...Originally Posted by jj451
I'm struggling with the following sum:
I've tried using a geometric series with the nth term equaling
but I don't think thats correct. I just checked in Wolfram Alpha and the answer there is mine multiplied by
Not really sure what to do really so any help would be appreciated.
This is a geometric series withand
, so applying the sum formula
we have
![\displaystyle \begin{align*} S_{N+1} &= \frac{1\left[1 - \left(\frac{1}{x}\right)^{N+2}\right]}{1 - \frac{1}{x}} \\ &= \frac{1 - \left(\frac{1}{x}\right)^{N+2}}{1 - \frac{1}{x}} \\ &= \frac{\frac{x^{N+2} - 1}{x^{N+2}}}{\frac{x - 1}{x}} \\ &= \frac{x\left(x^{N+2} - 1\right)}{x^{N+2}\left(x - 1\right)} \\ &= \frac{x^{N+2} - 1}{x^{N+1}\left(x - 1\right)} \end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} S_{N+1} &= \frac{1\left[1 - \left(\frac{1}{x}\right)^{N+2}\right]}{1 - \frac{1}{x}} \\ &= \frac{1 - \left(\frac{1}{x}\right)^{N+2}}{1 - \frac{1}{x}} \\ &= \frac{\frac{x^{N+2} - 1}{x^{N+2}}}{\frac{x - 1}{x}} \\ &= \frac{x\left(x^{N+2} - 1\right)}{x^{N+2}\left(x - 1\right)} \\ &= \frac{x^{N+2} - 1}{x^{N+1}\left(x - 1\right)} \end{align*})
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