The sum of 50 terms of the series $\frac{3}{1} $+ $\frac{5}{1+2^2} $+ $\frac{7}{1+2^2+3^2}$+ $\frac{11}{1+2^2+3^2+4^2}$... The numerator is increasing by 2.
The number of terms in denominator is increasing, and they are squares of natural numbers (i.e. $1^2, 2^2, 3^2, 4^2...$)
Options are:
(a)$\frac {100}{17}$
(b)$\frac {150}{17}$
(c)$\frac {200}{51}$
(d)$\frac {50}{17}$

Start by simplifying the denominator of the $n^{th}$ term.
$$U_n=\frac{2n+1}{\sum_{i=1}^ni^2}$$
$$=\frac{6(2n+1)}{n(n+1)(2n+1)}$$
$$=\frac{6}{n(n+1)}$$
$$=\frac{6}{n}-\frac{6}{n+1}$$
So the sum of the first 50 terms will be:
$$S_{50}=U_1+U_2+U_3+\cdots+U_{48}+U_{49}+U_{50}$$
$$=\frac{6}{1}-\frac{6}{2}+\frac{6}{2}-\frac{6}{3}+\frac{6}{3}-\frac{6}{4}+\cdots+\frac{6}{48}-\frac{6}{49}+\frac{6}{49}-\frac{6}{50}+\frac{6}{50}-\frac{6}{51}$$
$$=6-\frac{6}{51}$$
$$=\frac{100}{17}$$
