Question 931537
 1 power 4/ (1 x 3) + 2 power 4 /(3 x 5) +3 power 4/(5 x 7)………………n power 4/(2n-1)(2n+1) 
<pre>
{{{sum(k^4/((2k-1)(2k+1)),k=1,n)}}}

We break {{{k^4/((2k-1)(2k+1))}}}

into partial fractions, first by writing it as {{{(k^4+0k^3+0k^2+0k+0)/(4k^2+0k-1),

divide that out by long division, getting

{{{expr(1/4)x^2+expr(1/16)(1/(4k^2-1))}}}

Then break {{{1/((2k-1)(2k+1))}}} into partial fractions, getting

{{{expr(1/2)(1/(2k-1)-1/(2k+2))}}}

Then that's telescoping series.  Then use the sum of squares formula

and put it altogether and you end up with the expression for the sum as

{{{(n(n^3+2n^2+2n+1))/(6(2n+1))}}}

Edwin</pre>