Question 962542
<pre>
{{{1+a+a^2+""*""*""*""+a^(n-1)=(a^n-1)/(a-1)}}}

Prove true for n=1: (There is only one term)

{{{1=a^(1-1)}}}
{{{1=a^0}}}
{{{1=1}}}

So it is true for n=1

Assume it is true for all n &#8804; k

{{{1+a+a^2+""*""*""*""+a^(k-1)=(a^k-1)/(a-1)}}}

Add {{{a^k}}} to both sides

{{{1+a+a^2+""*""*""*""+a^(k-1)+a^k=(a^k-1)/(a-1)+a^k}}}

Simplifying the right side:

{{{(a^k-1)/(a-1)+a^k}}}

{{{(a^k-1)/(a-1)+a^k/1}}}

{{{(a^k-1)/(a-1)+a^k*(a-1)/(a-1)}}}

{{{((a^k-1)+a^k*(a-1))/(a-1)}}}

{{{(a^k-1+a^(k+1)-a^k)/(a-1)}}} 

{{{(cross(a^k)-1+a^(k+1)-cross(a^k))/(a-1)}}} 

{{{(-1+a^(k+1))/(a-1)}}}

{{{(a^(k+1)-1)/(a-1)}}}

And that is exactly what you would get if you substitute
n=k+1 in {{{(a^n-1)/(a-1)}}}.

So the formula is proved.

Edwin</pre>