Question

If $\sum _{r=1}^n{t}_r=\frac{1}{12}n(n+1)(n+2)$, then the value of $\sum _{r=1}^n\frac{1}{t_r}$ is

• A. $\frac{2n}{n+1}$
• B. $\frac{n}{(n+1)}$
• C. $\frac{4n}{n+1}$
• D. $\frac{3n}{n+1}$

Answer 1

The correct option is C $\frac{4n}{n+1}$

If we take $t_r=\frac{1}{4}n(n+1)$

$t_r=\frac{1}{4\times 3}n(n+1)\left[\right(n+2)-(n-1\left)\right]=\frac{1}{12}\left[n\right(n+1\left)\right(n+2)-n(n+1\left)\right(n-1\left)\right]t_1=\frac{1}{12}[\mathrm{1.2.3}-0]t_2=\frac{1}{12}[\mathrm{2.3.4}-\mathrm{1.2.3}]t_3=\frac{1}{12}[\mathrm{3.4.5}-\mathrm{2.3.4}]\underset{–}{t_n=\frac{1}{12}\left[n\right(n+1\left)\right(n+2)-n(n+1\left)\right(n-1\left)\right]}{\sum }_{r=1}^n{t}_r=\frac{1}{12}\left[n\right(n+1\left)\right(n+2\left)\right]Hence,t_r=\frac{n(n+1)}{4}\frac{1}{t_r}=\frac{4}{n(n+1)}=4\left[\frac{(n+1)-n}{n(n+1)}\right]=4[\frac{1}{n}-\frac{1}{(n+1)}]\frac{1}{t_1}=4[1-\frac{1}{2}]\frac{1}{t_2}=4[\frac{1}{2}-\frac{1}{3}]\underset{–}{\frac{1}{t_n}=4[\frac{1}{n}-\frac{1}{(n+1)}]}{\sum }_{r=1}^n\frac{1}{t_r}=4[1-\frac{1}{(n+1)}]=\frac{4n}{(n+1)}$