Question

The sum to $50$ terms of the series $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\dots$ is

• A. $\frac{100}{51}$
• B. $\frac{150}{17}$
• C. $\frac{200}{51}$
• D. $\frac{50}{17}$

Answer 1

The correct option is A $\frac{100}{51}$

$\begin{array}{c}\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\\ T_r=\frac{2r+1}{1^2+2^2+.......r^2}\\ =\frac{\left(2r+1\right)\times 2}{r\left(r+1\right)\left(2r+1\right)}\\ =\frac{2}{r\left(r+1\right)}\\ {\sum }_{r=1}^{50}{T}_r=2{\sum }_{r=1}^{50}\frac{\left(r+1\right)-r}{r\left(r+1\right)}=2{\sum }_{r=1}^{50}\left[\frac{1}{r}-\frac{1}{r+1}\right]\\ =2\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+........+\left(\frac{1}{50}-\frac{1}{51}\right)\right]\\ =2\left[1-\frac{1}{51}\right]\\ weget,\\ =\frac{100}{51}\\ Hence,\\ optionAiscorrectanswer.\end{array}$