Question

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

• A. 120
• B. 180
• C. 240
• D. 60

Answer 1

Answer: (A)

$t_n=\frac{2n+1}{\frac{n\left(n+1\right)\left(2n+1\right)}{6}}=\frac{6}{n\left(n+1\right)}=6\left(\frac{1}{n}-\frac{1}{n+1}\right)$
$S_n=6\left\{\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+.....-\frac{1}{21}\right\}=6\left\{\frac{1}{1}-\frac{1}{21}\right\}$
$=6\left(\frac{20}{21}\right)=\frac{120}{21}\Rightarrow k=120$