Question

If $a_k=\frac{1}{k(k+1)}$; then ${\left(\sum _{k=1}^n{a}_k\right)}^2$ is equal to

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

Answer 1

The correct option is A $\frac{n^2}{(n+1{)}^2}$

Simplifying

$\because a_k=\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$

Now, we have

$\therefore {\left(\sum _{k=1}^n{a}_K\right)}^2={\left[\sum _{k=1}^n(\frac{1}{k}-\frac{1}{k+1})\right]}^2$

$={[(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+......+(\frac{1}{n}-\frac{1}{n+1})]}^2$

$={(1-\frac{1}{n+1})}^2=\frac{n^2}{(n+1{)}^2}$