Question

If the sum $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+.....+$ upto $20$ terms is equal to $\frac{k}{21}$, then $k$ is equal to

• A. $120$
• B. $180$
• C. $240$
• D. $60$

Answer 1

The correct option is A $120$

$\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+.....+$ upto $20$ terms $=\sum _{r=1}^{20}\frac{2r+1}{\sum r^2}=\sum _{r=1}^{20}\frac{(2r+1)}{\frac{r(r+1)(2r+1)}{6}}$

$=6\sum _{r=1}^{20}\frac{r+1-r}{r(r+1)}=6\sum _{r=1}^{20}(\frac{1}{r}-\frac{1}{r+1})=6(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots -\frac{1}{21})$

$=6(\frac{21-1}{21})=\frac{120}{21}$

$⟹\frac{k}{21}=\frac{120}{21}⟹k=120$