Question

If $a_k=\frac{1}{K(K+1)}$ for $K=1,2,3,.....n,$ then ${\left({\sum }_{k=1}^n{a}_k\right)}^2=$

• A. $\frac{n}{n+1}$
• B. $\frac{n^2}{(n+1{)}^2}$
• C. $\frac{n^4}{(n+1{)}^4}$
• D. $\frac{n^6}{(n+1{)}^6}$

Answer 1

Answer: (B)

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

$\begin{array}{c}{a}_k=\frac{1}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\\ {\left({\sum }_{k=1}^n{a}_x\right)}^2\\ {\left({\sum }_{k=1}^n\left(\frac{1}{k}-\frac{1}{k+1}\right)\right)}^2\\ =\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)\left(\frac{1}{n}\frac{-1}{n+1}\right)\\ ={\left(1-\frac{1}{n+1}\right)}^2\\ ={\left(\frac{n+1-1}{n+1}\right)}^2\\ =\frac{n^2}{{\left(n+1\right)}^2}\end{array}$

Hence, this is the answer.