1. ## Fractions and indices

this is a problem from a previous years MAOL math competition Matemaattisten Aineiden Opettajien Liitto MAOL ry : Etusivu_ala

$\frac{1}{2002^{-10}}+\frac{1}{2002^{-9}}\mbox{...}+\frac{1}{2002^0}\mbox{...}+\frac{1}{ 2002^{9}}+\frac{1}{2002^{10}}$

Could someone please help me solve this? I tried various rules for indices, but noticed that the ones we have been taught work only for multiplication and division.

2. Originally Posted by Coach
this is a problem from a previous years MAOL math competition Matemaattisten Aineiden Opettajien Liitto MAOL ry : Etusivu_ala

$\frac{1}{2002^{-10}}+\frac{1}{2002^{-9}}\mbox{...}+\frac{1}{2002^0}\mbox{...}+\frac{1}{ 2002^{9}}+\frac{1}{2002^{10}}$

Could someone please help me solve this? I tried various rules for indices, but noticed that the ones we have been taught work only for multiplication and division.

$\frac{1}{2002^{-10}}+\frac{1}{2002^{-9}}\mbox{...}+\frac{1}{2002^0}\mbox{...}+\frac{1}{ 2002^{9}}+\frac{1}{2002^{10}}$

....... $
\ \ \ \ =\frac{1}{2002^{-10}}\left[ 1+\frac{1}{2002} + ... + \frac{1}{2002^{20}} \right]
$

The expression inside the square brackets is a finite geometric series whose sum is:

$
\left[ 1+\frac{1}{2002} + ... + \frac{1}{2002^{20}} \right]= \frac{1-2002^{-21}}{1-2002^{-1}}
$

So:

$\frac{1}{2002^{-10}}+\frac{1}{2002^{-9}}\mbox{...}+\frac{1}{2002^0}\mbox{...}+\frac{1}{ 2002^{9}}+\frac{1}{2002^{10}}$

....... $
\ \ \ \ =\frac{1}{2002^{-10}}\left[ 1+\frac{1}{2002} + ... + \frac{1}{2002^{20}} \right]
$

....... $
\ \ \ \ =2002^{10} \frac{1-2002^{-21}}{1-2002^{-1}}
$

Which you can simplify further if required

RonL

3. The terms form a geometric progression with ratio $\displaystyle\frac{1}{2002}$ and the sum contain 21 terms.