Question

What is the sum of n terms of the series `1^4/(1*3) +2^4/(3*5) +3^4/(5*7) +cdots+n^4/[(2n-1)(2n+1)` ?

Answer 1

See below.

Explanation:

We have

`n^4 = (4n^2-1) q(n)+r(n)`

with

`q(n) = c_1n^2+c_2n+c_2` and
`r(n) = an+b`

solving for `c_1,c_2,c_3,a,b` we get

`q(n) = 1/4(n^2+1/4)` and
`r(n) = 1/16`

and then

`n^4/(4n^2-1) = 1/4(n^2+1/4) + 1/16 1/((2n-1)(2n+1))` and then

`sum_(n=0)^m n^4/(4n^2-1) = m/16+1/4 sum_(n=0)^m n^2+1/32(sum_(n=0)^m 1/(2n-1)-sum_(n=0)^m 1/(2n+1))`

but

`sum_(n=0)^m 1/(2n-1)-sum_(n=0)^m 1/(2n+1) = 1-1/(2m+1)` then

`sum_(n=0)^m n^4/(4n^2-1) = m/16+1/4 sum_(n=0)^m n^2+1/32(1-1/(2m+1))`

We left the concluding results to the reader as an exercise.