Let for $$n = 1, 2, \ldots, 50$$, $$S_n$$ be the sum of the infinite geometric progression whose first term is $$n^2$$ and whose common ratio is $$\frac{1}{(n+1)^2}$$. Then the value of $$\frac{1}{26} + \sum_{n=1}^{50} \left(S_n + \frac{2}{n+1} - n - 1\right)$$ is equal to

We need to find $$\frac{1}{26} + \sum_{n=1}^{50}\left(S_n + \frac{2}{n+1} - n - 1\right)$$ where $$S_n$$ is the sum of the infinite GP with first term $$n^2$$ and common ratio $$\frac{1}{(n+1)^2}$$.

Find $$S_n$$:

$$S_n = \frac{n^2}{1 - \frac{1}{(n+1)^2}} = \frac{n^2 \cdot (n+1)^2}{(n+1)^2 - 1} = \frac{n^2(n+1)^2}{n(n+2)} = \frac{n(n+1)^2}{n+2}$$

Simplify $$S_n$$ by polynomial division:

Dividing $$n(n+1)^2 = n^3 + 2n^2 + n$$ by $$(n+2)$$:

$$n^3 + 2n^2 + n = (n+2) \cdot n^2 + n$$

So $$\frac{n(n+1)^2}{n+2} = n^2 + \frac{n}{n+2} = n^2 + 1 - \frac{2}{n+2}$$

Substitute into the summand:

$$S_n + \frac{2}{n+1} - n - 1 = n^2 + 1 - \frac{2}{n+2} + \frac{2}{n+1} - n - 1$$

$$= n^2 - n + \frac{2}{n+1} - \frac{2}{n+2}$$

Sum from $$n = 1$$ to $$50$$:

$$\sum_{n=1}^{50}\left(n^2 - n\right) + 2\sum_{n=1}^{50}\left(\frac{1}{n+1} - \frac{1}{n+2}\right)$$

For the first sum:

$$\sum_{n=1}^{50}n(n-1) = \sum_{n=1}^{50}n^2 - \sum_{n=1}^{50}n = \frac{50 \cdot 51 \cdot 101}{6} - \frac{50 \cdot 51}{2}$$

$$= \frac{50 \cdot 51}{6}(101 - 3) = \frac{50 \cdot 51 \cdot 98}{6} = \frac{249900}{6} = 41650$$

For the second sum (telescoping):

$$2\left(\frac{1}{2} - \frac{1}{52}\right) = 2 \cdot \frac{52 - 2}{104} = 2 \cdot \frac{50}{104} = \frac{100}{104} = \frac{25}{26}$$

Compute the final answer:

$$\frac{1}{26} + 41650 + \frac{25}{26} = \frac{1}{26} + \frac{25}{26} + 41650 = 1 + 41650 = 41651$$

The answer is $$\boxed{41651}$$.