Question

If $\sum _{k=1}^{n}f\left(k\right)=\frac{n}{2(n+2)}$, then the value of $\sum _{k=1}^{10}\frac{1}{f\left(k\right)}$ is equal to

• A. $570$
• B. $480$
• C. $560$
• D. $550$

Answer 1

The correct option is A $570$

Let $S_n=\sum _{k=1}^{n}f\left(k\right)=\frac{n}{2(n+2)}$
$S_n-S_{n-1}=\frac{n}{2(n+2)}-\frac{n-1}{2(n+1)}\Rightarrow S_n-S_{n-1}=\frac{1}{(n+1)(n+2)}\Rightarrow f\left(n\right)=\frac{1}{(n+1)(n+2)}\Rightarrow \frac{1}{f\left(n\right)}=(n+1)(n+2)\Rightarrow \frac{1}{f\left(n\right)}=n^2+3n+2\sum _{k=1}^{10}\frac{1}{f\left(k\right)}=\sum _{k=1}^{10}(k^2+3k+2)=\sum _{k=1}^{10}{k}^2+3\sum _{k=1}^{10}k+20=\frac{10\left(11\right)\left(21\right)}{6}+3\frac{10\left(11\right)}{2}+20=385+165+20=570$