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Question Number 91328 by Maclaurin Stickker last updated on 30/Apr/20

$$Findthesum$$$$\frac{9}{1+\sqrt{2}}+\frac{10-\lfloor \sqrt{2}\rfloor }{\sqrt{2}+\sqrt{3}}+\frac{10-\lfloor \sqrt{3}\rfloor }{\sqrt{3}+\sqrt{4}}+...+\frac{10-\lfloor \sqrt{99}\rfloor }{\sqrt{99}+\sqrt{100}}$$$$\lfloor x\rfloor =thegreatestintegerfunction$$

Commented by Prithwish Sen 1 last updated on 30/Apr/20

$$\lfloor \sqrt{1}\rfloor \mathrm{to}\lfloor \sqrt{3}\rfloor \backsim 1$$$$\lfloor \sqrt{4}\rfloor \mathrm{to}\lfloor \sqrt{8}\rfloor \backsim 2$$$$\lfloor \sqrt{9}\rfloor \mathrm{to}\lfloor \sqrt{15}\rfloor \backsim 3$$$$:$$$$:$$$$\lfloor \sqrt{81}\rfloor \mathrm{to}\lfloor \sqrt{99}\rfloor \backsim 9$$$$\therefore \mathrm{the}\mathrm{series}\mathrm{becomes}$$$$9(\frac{1}{\sqrt{1}+\sqrt{2}}+..+\frac{1}{\sqrt{3}+\sqrt{4}})+8(\frac{1}{\sqrt{4}+\sqrt{5}}+..+\frac{1}{\sqrt{8}+\sqrt{9}})$$$$+7(\frac{1}{\sqrt{9}+\sqrt{10}}+..+\frac{1}{\sqrt{15}+\sqrt{16}})+......2(\frac{1}{\sqrt{64}+\sqrt{65}}+..$$$$\frac{1}{\sqrt{80}+\sqrt{81}})+1(\frac{1}{\sqrt{81}+\sqrt{82}}+..+\frac{1}{\sqrt{99}+\sqrt{100}})$$$$=9(\sqrt{4}-1)+8(\sqrt{9}-\sqrt{4})+7(\sqrt{16}-\sqrt{9})+....+2(\sqrt{81}-\sqrt{64})$$$$+1(\sqrt{100}-\sqrt{81})$$$$=9+8+7+6+5+4+3+2+1$$$$=45\mathbf{please}\mathbf{check}.$$

Commented by Prithwish Sen 1 last updated on 30/Apr/20

$$35+3\sqrt{11}$$

Commented by mr W last updated on 30/Apr/20

$$nicesolution!$$

Commented by Prithwish Sen 1 last updated on 30/Apr/20

$$\mathrm{am}\mathrm{I}\mathrm{wrong}\mathrm{Sir}?$$

Commented by Prithwish Sen 1 last updated on 30/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}!\mathrm{I}\mathrm{fix}\mathrm{it}.$$

Commented by Prithwish Sen 1 last updated on 30/Apr/20

$$\mathrm{Thanks}\mathrm{Sir}!$$

Commented by Maclaurin Stickker last updated on 30/Apr/20

$$theansweris135$$

Commented by mr W last updated on 30/Apr/20

$$iftheansweris135,thenyourquestion$$$$iswrong.pleaserecheck!$$

Commented by Maclaurin Stickker last updated on 30/Apr/20

$$Ithinkthebookiswrong.$$$$Iwillchecktheoriginalquestion.$$

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