Question 941989
We first want to prove that 
if equation (1) were true for some value of n,


(1)  {{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n)

))}}}{{{""=""}}}{{{n(n+1)/(2(2n+1))}}} then it would also be true when we replace n by n+1.  That is,
we want to show that if (1) holds for some value of n, then we 
can erase the ??? below and replace it with an equal sign:

{{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",red(n+1))

))}}}{{{"???"}}}{{{red((n+1)^"")(red((n+1)^"")+1)/(2(2red((n+1)^"")+1))}}}

which simplifies to:

(2)  {{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n+1)

))}}}{{{"???"}}}{{{(n+1)(n+2)/(2(2n+3))}}}

-----------------------

So we add the (n+1)st term to both sides of (1)

{{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n)

))}}}{{{""+""}}} {{{(n+1)^2/((2(n+1)^""-1)(2(n+1)^""+1))}}} {{{""=""}}} {{{n(n+1)^""/(2(2n+1)^"")}}} {{{""+""}}}  {{{(n+1)^2/((2(n+1)^""-1)(2(n+1)^""+1))}}}

and simplify.  The left side is just the sum to n+1 terms instead of n
terms:

{{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n+1)

))}}} {{{""=""}}} {{{n(n+1)^""/(2(2n+1)^"")}}} {{{""+""}}}  {{{(n+1)^2/(2n+1)(2n+3))}}}

Get an LCD by multiplying the first fraction on the right by {{{((2n+3))/((2n+3))}}} and the second by {{{2/2}}}

{{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n+1)

))}}} {{{""=""}}} {{{n(n+1)^""(2n+3)/(2(2n+1)^""(2n+3))}}} {{{""+""}}}  {{{2(n+1)^2/2(2n+1)(2n+3))}}}

{{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n+1)

))}}} {{{""=""}}} {{{n(n+1)^""(2n+3)/(2(2n+1)^""(2n+3))}}} {{{""+""}}}  {{{2(n+1)^2/2(2n+1)(2n+3))}}}

Combine numerators over the LCD

{{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n+1)

))}}} {{{""=""}}} {{{(n(n+1)^""(2n+3)+2(n+1)^2)/(2(2n+1)^""(2n+3))}}}

Factor out (n+1) on top

{{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n+1)

))}}} {{{""=""}}} {{{((n+1)(n(2n+3)+2(n+1)^""))/(2(2n+1)^""(2n+3))}}}

{{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n+1)

))}}} {{{""=""}}} {{{((n+1)(2n^2+3n+2n+2))/(2(2n+1)^""(2n+3))}}}

{{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n+1)

))}}} {{{""=""}}} {{{((n+1)(2n^2+5n+2))/(2(2n+1)^""(2n+3))}}}

{{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n+1)

))}}} {{{""=""}}} {{{((n+1)(2n+1)(n+2)^"")/(2(2n+1)^""(2n+3))}}}

Now cancel the (2n+1)'s

{{{sum(r^2/((2r-1)(2r+1)),

matrix(1,3,"","",r=1),

matrix(1,4,"","","",n+1)

))}}} {{{""=""}}} {{{((n+1)(n+2)^"")/(2(2n+3))}}}

That is the same as (2)

All that's left is to show that (1) is true for n=1


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

{{{1/((2-1)(2+1)        )}}}{{{""=""}}}{{{1(2)/(2(2+1))}}}

{{{1/((1)(3)        )}}}{{{""=""}}}{{{2/(2(3))}}}

{{{1/3}}} {{{""=""}}} {{{1/3}}}

So since it works for n=1, it works for n=2. Then
since it works for n=2, it works for n=3, then for n=4 
and so on forever for all values on n.

Edwin</pre>