Question

If $\sum _{r=1}^n{t}_r=\sum _{k=1}^n\sum _{j=1}^k\sum _{i=1}^j\left(2\right),$ then $\sum _{r=1}^n\frac{1}{t_r}$ is

• A. $\frac{1}{n+1}$
• B. $\frac{n}{n+1}$
• C. $\frac{n+1}{n}$
• D. $n+1$

Answer 1

The correct option is B $\frac{n}{n+1}$

$\sum _{r=1}^n{t}_r=\sum _{k=1}^n\sum _{j=1}^k\sum _{i=1}^j\left(2\right)=\sum _{k=1}^n\sum _{j=1}^{k}2j=\sum _{k=1}^{n}k(1+k)\therefore t_r=r(1+r)\Rightarrow \frac{1}{t_r}=\frac{1}{r(r+1)}=\frac{1}{r}-\frac{1}{r+1}$

Now, $\sum _{r=1}^n\frac{1}{t_r}$
$=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+⋮⋮+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}=\frac{n}{n+1}$