# Thread: From a series how do we get this result?

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2. ## Re: From a series how do we get this result?

Originally Posted by math951
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This all I can offer you:
$\displaystyle \sum\limits_{k = 1}^x {{{\left( {\frac{5}{6}} \right)}^{k - 1}}} = \frac{{1 - {{\left( {\tfrac{5}{6}} \right)}^x}}}{{1 - \left( {\tfrac{5}{6}} \right)}}$

Because it $|r|<1$ then $\displaystyle \sum\limits_{k = 1}^x {{r^{k - 1}}} = \frac{{1 - {r^x}}}{{1 - r}}$

3. ## Re: From a series how do we get this result?

If you want a little more detail, let $\displaystyle S(n)= 1+ r+ r^2+ \cdot\cdot\cdot+ r^n$. $\displaystyle S(n)= 1+ (r+ r^2+ \cdot\cdot\cdot+ r^n)= 1+ r(1+ r+ \cdot\cdot\cdot+ r^{n-1})$.

Add and subtract $\displaystyle r^n$ inside the parentheses:
$\displaystyle S(n)= 1+ r(1+ r+ \cdot\cdot\cdot+ r^{n-1}+ r^n- r^n)= 1+ r(1+ \cdot\cdot\cdot+ r^n)- r^{n+1}$
$\displaystyle S(n)= 1+ rS(n)- r^{n+1}$
$\displaystyle S(n)- rS(n)= (1- r)S(n)= 1- r^{n+1}$
$\displaystyle S(n)= \frac{1- r^{n+1}}{1- r}$.

Here $\displaystyle r= \frac{5}{6}$.