Hello all. Can somebody please help with the intermediate steps here? I have reached the point given, but not sure how to advance to the final solution. Many thanks

$\displaystyle \sum_{i=0}^9 (1/2z)^{n}=\frac{1-(2z)^{-10}}{1-(2z)^{-1}}$

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- Mar 1st 2010, 10:00 AMjayd1988Help finding closed form of sum
Hello all. Can somebody please help with the intermediate steps here? I have reached the point given, but not sure how to advance to the final solution. Many thanks

$\displaystyle \sum_{i=0}^9 (1/2z)^{n}=\frac{1-(2z)^{-10}}{1-(2z)^{-1}}$ - Mar 1st 2010, 10:35 AMKrizalid
that's all, that's a closed form.

don't know what you want to do further. - Mar 1st 2010, 10:42 AMjayd1988
Sorry, I wanted to know the necessary steps I need to take to get it into the closed form, the first equation is where I got to, I have the solution but don't know how to get there :-)

- Mar 1st 2010, 10:43 AMKrizalid
ah okay then, can you prove that $\displaystyle \sum_{j=0}^n a^j=\frac{a^{n+1}-1}{a-1}$ ?

if you can do that, then you're done. (This is a fact known as geometric sum.) - Mar 1st 2010, 10:48 AMjayd1988
Many thanks, solved.