Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 123108 by benjo_mathlover last updated on 23/Nov/20

Commented by benjo_mathlover last updated on 23/Nov/20

$$\frac{1}{\sqrt{2}+2}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\frac{1}{4\sqrt{5}+5\sqrt{4}}+$$$$...+\frac{1}{4012008\sqrt{4012009}+4012009\sqrt{4012008}}?$$

Commented by liberty last updated on 23/Nov/20

$$weareaskedtocompute\underset{k=1}{\sum ^{4012008}}\frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}$$$$considertheterm\frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}=\frac{1}{\sqrt{k}\sqrt{k+1}(\sqrt{k}+\sqrt{k+1})}$$$$=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$$$$sotheequationbecomes$$$$\underset{k=1}{\sum ^{4012008}}(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}})=1-\frac{1}{\sqrt{4012009}}$$$$=1-\frac{1}{\sqrt{{2003}^2}}=\frac{2002}{2003}.▴$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com