Question 826711
Show that in any GP,
{{{S[2n]}}}{{{":"}}}{{{S[n]}}}{{{""=""}}}{{{r^n+1}}}{{{":"}}}{{{1}}}   
<pre>
{{{S[n]}}}{{{""=""}}}{{{a[1](1-r^n)/(1-r)}}}

So

{{{S[2n]}}}{{{""=""}}}{{{a[1](1-r^(2n))/(1-r)}}}

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{{{S[2n]/S[n]}}}{{{""=""}}}{{{(a[1](1-r^(2n))/(1-r))/(a[1](1-r^n)/(1-r))}}}{{{""=""}}}{{{a[1](1-r^(2n))/(1-r)}}}{{{"÷"}}}{{{a[1](1-r^n)/(1-r)}}}{{{""=""}}}{{{a[1](1-r^(2n))/(1-r)}}}{{{""*""}}}{{{(1-r)/a[1](1-r^n)}}}{{{""=""}}}
{{{cross(a[1])(1-r^(2n))/(cross(1-r))}}}{{{""*""}}}{{{(cross(1-r))/cross(a[1])(1-r^n)}}}{{{""=""}}}{{{(1-r^(2n))/(1-r^n)}}}{{{""=""}}}{{{((1-r^n)(1+r^n))/(1-r^n)}}}{{{""=""}}}{{{((cross(1-r^n))(1+r^n))/(cross(1-r^n))}}}{{{""=""}}}
{{{1+r^n}}}{{{""=""}}}{{{(r^n+1)/1}}}

{{{S[2n]/S[n]}}}{{{""=""}}}{{{(r^n+1)/1}}}

is equivalent to the proportion:

{{{S[2n]}}}{{{":"}}}{{{S[n]}}}{{{""=""}}}{{{r^n+1}}}{{{":"}}}{{{1}}} 

Edwin</pre>