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Moo · #4 November 13th, 2008, 12:55 PM

Hello,

Quote:

Originally Posted by tttcomrader

Thanks for your help, but may I ask what formula did we use here?

$\frac{X^k-z^k}{X-z}=\frac{X^k \left(1-\left(\tfrac zX\right)^k\right)}{X \left(1-\frac zX\right)}$

$=X^{k-1} \times \frac{1-\left(\tfrac zX\right)^k}{1-\frac zX}$

You can recognize that this is the sum of a geometric sequence :
$\sum_{i=0}^{k-1} \left(\tfrac zX\right)^i=\frac{1-\left(\tfrac zX\right)^k}{1-\frac zX}$

Hence
$\frac{X^k-z^k}{X-z}=X^{k-1} \sum_{i=0}^{k-1} \left(\tfrac zX\right)^i=\sum_{i=0}^{k-1} z^i \cdot X^{(k-1)-i}$

and this gives the formula Laurent wrote.

The following users thank Moo for this useful post:
tttcomrader
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