It's easier to deal with the symbolic sum (in sigma notation),
[tex]\displaystyle\sum_{k=1}^{50}\frac{(2k+1)^2+1}{(2k+1)^2-1}[/tex]
Expanding the terms in the fraction, computing the quotient, and decomposing into partial fractions gives
[tex]\dfrac{(2k+1)^2+1}{(2k+1)^2-1} = \dfrac{4k^2 + 4k + 2}{4k^2 + 4k}[/tex]
[tex]=\dfrac12\times\dfrac{2k^2 + 2k + 1}{k^2 + k}[/tex]
[tex]=\dfrac12\left(2+\dfrac1{k(k+1)}\right)[/tex]
[tex]=\dfrac12\left(2 + \dfrac1k - \dfrac1{k+1}\right)[/tex]
and it's the latter two terms that reveal a telescoping pattern.
In case you need more details about the partial fraction decomposition, we are looking for coefficients a and b such that
[tex]\dfrac1{k(k+1)}=\dfrac ak+\dfrac b{k+1}[/tex]
or
[tex]1 = a(k+1) +bk =(a+b)k+a[/tex]
which gives a = 1, and a + b = 0 so that b = -1.
Our sum has been rearranged as
[tex]\displaystyle\frac12\sum_{k=1}^{50}\left(2+\frac1k-\frac1{k+1}\right)=\sum_{k=1}^{50}1+\frac12\sum_{k=1}^{50}\left(\frac1k-\frac1{k+1}\right)=50+\frac12\sum_{k=1}^{50}\left(\frac1k-\frac1{k+1}\right)[/tex]
The remaining telescoping sum is
1/2 [(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + … + (1/48 - 1/49) + (1/49 - 1/50) + (1/50 - 1/51)]
and you can see how there are pairs of numbers that cancel, so that the sum reduces to
1/2 [1/1 - 1/51] = 1/2 [1 - 1/51] = 1/2 × 50/51 = 25/51
So, our original sum ends up being
[tex]\displaystyle\sum_{k=1}^{50}\frac{(2k+1)^2+1}{(2k+1)^2-1} = 50 + \frac{25}{51} = \boxed{\dfrac{2575}{51}}[/tex]
