Question

The value of $S=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1++\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1++\frac{1}{{\left(2014\right)}^2}+\frac{1}{{\left(2015\right)}^2}}$ is

• A. $2015$
• B. $2015-\frac{1}{2015}$
• C. $2016-\frac{1}{2016}$
• D. $2014-\frac{1}{2014}$

Answer 1

Answer: (B)

$2015-\frac{1}{2015}$

The first term $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}=\sqrt{1+1+\frac{1}{4}}=\sqrt{\frac{9}{4}}=\frac{3}{2}=1+\frac{1}{2}=1+1-\frac{1}{2}$

The second term $\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=\sqrt{1+\frac{1}{4}+\frac{1}{9}}=\sqrt{\frac{49}{36}}=\frac{7}{6}=1+\frac{1}{6}=1+\frac{1}{2}-\frac{1}{3}$

The Third term similarily can be written as $1+\frac{1}{3}-\frac{1}{4}$

Hence, $S=(1+1-\frac{1}{2})+(1+\frac{1}{2}-\frac{1}{3})+...+(1+\frac{1}{2014}-\frac{1}{2015})=2015-\frac{1}{2015}$