How to calculate summation of $\frac{n^2 + n + 1 }{n^2 + n}$ over $1$ to $25$? Can we calculate the summation of numerator, and denominator and then divide them both? But that does not seems to be a plausible way. So, I reduced it to $S = 1+ \frac{1}{n^2+n}$. But now I do not how to compute this sum.

Using telescoping sum and the simplification you found,
$$\\\sum_1^{25} \frac{n^2+n+1}{n^2+n}\\
= \sum_1^{25}\left(1+\frac{1}{n(n+1)}\right)\\
= \sum_1^{25}1 + \sum_1^{25}\frac{1}{n(n+1)}\\
= 25 + \sum_1^{25}\left(\frac{1}{n}-\frac{1}{n+1}\right)\\
= 25 + \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{25}-\frac{1}{26}\\
=25\frac{25}{26}$$
