Question 1116740
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The formula in your post is written &nbsp;&nbsp;<U>I N C O R R E C T L Y</U>.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The correct formula is &nbsp;&nbsp;<U>T H I S</U>:


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{ax/(a-x)}}} = {{{x}}} + {{{x^2/a}}} + {{{x^3/a^2}}} + {{{x^4/a^3}}} + &nbsp;. &nbsp;. &nbsp;. &nbsp;,  &nbsp;&nbsp;ect., &nbsp;&nbsp;to infinity.


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It is not difficult to prove it  &nbsp;&nbsp;//  &nbsp;&nbsp;if you know the formula for the sum of an infinite geometric series . . . 


<pre>
{{{a/(a-x)}}} = {{{a*(1/(a*(1-x/a))))}}} = {{{1/(1-x/a)}}}.


The very right fraction  {{{1/(1-x/a)}}}  is the sum of an infinite geometric progression with the first term of "1" and the common ratio of {{{x/a}}}:


{{{1/(1-x/a)}}} =  1 + {{{x/a}}} + {{{x^2/a^2}}} + {{{x^3/a^3}}} + . . .      (1)


Now multiply both sides of (1) by x, and you will get the required solution


{{{(ax)/(a-x)}}} = x + {{{x^2/a}}} + {{{x^3/a^2}}} + {{{x^4/a^3}}} + . . .


QED.
</pre>

Solved.