Question

Find the sum of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+....n\text{terms}$

Answer 1

$S={\sum }_{r=1}^n\sqrt{1+\frac{1}{r^2}+\frac{1}{(r+1{)}^2}}$

$={\sum }_{r=1}^n\sqrt{\frac{\left(r\right(r+1){)}^2+r^2+(r+1{)}^2}{\left(r\right(r+1){)}^2}}{\sum }_{r=0}^n\frac{\sqrt{r^4+2r^3+3r^2+2r+1}}{r(r+1)}$

$\Rightarrow S={\sum }_{r=1}^n\frac{\sqrt{(r^2+r+1{)}^2}}{r(r+1)}={\sum }_{r=1}^n\frac{r^2+r+1}{r^2+r}={\sum }_{r=1}^n(1+\frac{1}{r(r+1)})={\sum }_{r=1}^{n}1+{\sum }_{r=1}^n(\frac{1}{r}-\frac{1}{r+1})$

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