Find-the-sum-9-1-2-10-2-2-3-10-3-3-4-10-99-99-100-x-the-greatest-integer-function-

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

$${Find}\:{the}\:{sum} \\ $$$$\frac{\mathrm{9}}{\mathrm{1}+\sqrt{\mathrm{2}}}+\frac{\mathrm{10}−\lfloor\sqrt{\mathrm{2}}\rfloor}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}}+\frac{\mathrm{10}−\lfloor\sqrt{\mathrm{3}}\rfloor}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}}+…+\frac{\mathrm{10}−\lfloor\sqrt{\mathrm{99}}\rfloor}{\:\sqrt{\mathrm{99}}+\sqrt{\mathrm{100}}} \\ $$$$\lfloor{x}\rfloor={the}\:{greatest}\:{integer}\:{function} \\ $$

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

⌊(√1)⌋ to ⌊(√3)⌋ ∽ 1 ⌊(√4)⌋ to ⌊(√8)⌋ ∽ 2 ⌊(√9)⌋ to ⌊(√(15))⌋∽ 3 : : ⌊(√(81))⌋ to ⌊(√(99))⌋ ∽ 9 ∴ the series becomes 9((1/( (√1)+(√2))) +..+(1/( (√3)+(√4))))+8((1/( (√4)+(√5)))+..+(1/( (√8)+(√9)))) +7((1/( (√9)+(√(10))))+..+(1/( (√(15))+(√(16))))) +......2((1/( (√(64))+(√(65))))+.. (1/( (√(80))+(√(81)))))+1((1/( (√(81))+(√(82)))) +..+(1/( (√(99))+(√(100))))) =9((√4)−1)+8((√9)−(√4))+7((√(16))−(√9))+....+2((√(81))−(√(64))) +1((√(100))−(√(81))) =9+8+7+6+5+4+3+2+1 =45 please check.

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

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