Question

$a_k=\frac{1}{k(k+1)}$ for $k=1,2,3,\dots nthen(\sum _{k=1}^n{a}_k{)}^2$

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

Answer 1

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

$a_k=\frac{1}{\left(k\right)}\times \frac{1}{(k+1)}$
$=\frac{(k+1)-k}{\left(k\right)(k+1)}=\frac{1}{k}-1/k+1$
So, the value of $ak=\frac{1}{k}-1/k+1$
So ${\sum }_1^{1}ak=(1-1/2)+(1/2-1/3)$
$+.........(\frac{1}{n-1}-\frac{1}{n})+(\frac{1}{n}-\frac{1}{n+1})$
$=1-\frac{1}{n+1}=\frac{n}{n+1}$
${\sum }_1^{1}ak=\frac{n}{n+1}$
$({\sum }_1^{1}ak{)}^2=\frac{n^2}{(n+1{)}^2}$