Question

Evaluate $\sum _{r=1}^{998}\sqrt{1+\frac{1}{{\left(r+1\right)}^2}+\frac{1}{{\left(r+2\right)}^2}}$

Answer 1

$T_r=\sqrt{1+\frac{1}{(r+1{)}^2}+\frac{1}{(r+2{)}^2}}$

$T_r=\sqrt{\frac{(r+1{)}^2(r+2{)}^2+(r+1{)}^2+(r+2{)}^2}{(r+1{)}^2(r+2{)}^2}}$

$T_r=\sqrt{\frac{r^4+6r^3+15r^2+18r+9}{(r+1{)}^2(r+2{)}^2}}$

$T_r=\sqrt{\frac{(r^2+3r+3{)}^2}{(r+1{)}^2(r+2{)}^2}}$

$T_r=\frac{(r^2+3r+3)}{(r+1)(r+2)}$

$T_r=\frac{r^2+3r+2+1}{(r+1)(r+2)}$

$T_r=1+\frac{1}{(r+1)(r+2)}$

$T_r=1+\frac{1}{r+1}-\frac{1}{r+2}$

$S_{998}={\sum }_{n=1}^{998}(1+\frac{1}{r+1}-\frac{1}{r+2})$

$T_1=1+\frac{1}{2}-\frac{1}{3}$

$T_2=1+\frac{1}{3}-\frac{1}{4}$

.

.

$T_{998}=1+\frac{1}{999}-\frac{1}{1000}$

Adding these we get,

$S_{998}=(1+1+1+\mathrm{1.....998}terms)+\frac{1}{2}-\frac{1}{1000}$

$S_{998}=998-\frac{998}{1000}$

$S_{998}=\frac{998\ast 999}{1000}$