Question

Sum of the series $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+......$ upto 100 terms is

• A. $\frac{100}{0.06}$
• B. $\frac{600}{101}$
• C. $\frac{601}{101}$
• D. $\frac{101}{600}$

Answer 1

The correct option is B $\frac{600}{101}$

$\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+......100terms{T}_r=\frac{(2r+1)}{\frac{r(r+1)(2r+1)}{6}}=\frac{6}{r(r+1)}=6[\frac{1}{r}-\frac{1}{r+1}]\therefore S_{100}=6[\frac{1}{1}-\frac{1}{101}]=\frac{600}{101}$