Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 159891 by tounghoungko last updated on 22/Nov/21

$$S=\underset{k=1}{\sum ^{2002}}\sqrt{\frac{k^2+1}{k^2}+\frac{1}{{(k+1)}^2}}=?$$

Answered by chhaythean last updated on 22/Nov/21

$$\mathrm{S}=\underset{\mathrm{k}=1}{\sum ^{2002}}\sqrt{\frac{{\mathrm{k}}^2+1}{{\mathrm{k}}^2}+\frac{1}{{(\mathrm{k}+1)}^2}}$$$$\mathrm{Notice}\mathrm{that}:\frac{{\mathrm{k}}^2+1}{{\mathrm{k}}^2}+\frac{1}{{(\mathrm{k}+1)}^2}$$$$=\frac{({\mathrm{k}}^2+1){(\mathrm{k}+1)}^2+{\mathrm{k}}^2}{{\mathrm{k}}^2{(\mathrm{k}+1)}^2}$$$$=\frac{{\mathrm{k}}^4+{2\mathrm{k}}^3+{\mathrm{k}}^2+{\mathrm{k}}^2+2\mathrm{k}+1+{\mathrm{k}}^2}{{\mathrm{k}}^2{(\mathrm{k}+1)}^2}$$$$=\frac{{\mathrm{k}}^2{(\mathrm{k}+1)}^2+2\mathrm{k}(\mathrm{k}+1)+1}{{\mathrm{k}}^2{(\mathrm{k}+1)}^2}$$$$=\frac{{[\mathrm{k}(\mathrm{k}+1)+1]}^2}{{\mathrm{k}}^2{(\mathrm{k}+1)}^2}$$$$\Rightarrow \sqrt{\frac{{\mathrm{k}}^2+1}{{\mathrm{k}}^2}+\frac{1}{{(\mathrm{k}+1)}^2}}=\frac{\mathrm{k}(\mathrm{k}+1)+1}{\mathrm{k}(\mathrm{k}+1)}=1+\frac{1}{\mathrm{k}}-\frac{1}{(\mathrm{k}+1)}$$$$\mathrm{therefore}:\mathrm{S}=\underset{\mathrm{k}=1}{\sum ^{2002}}1+\underset{\mathrm{k}=1}{\sum ^{2002}}(\frac{1}{\mathrm{k}}-\frac{1}{\mathrm{k}+1})$$$$=2002+1-\frac{1}{2003}$$$$\mathrm{So}\overline{)\begin{array}{c}\mathrm{S}=\frac{4012008}{2003}\end{array}}$$

Commented by tounghoungko last updated on 22/Nov/21

$$yestelescopicseries$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com