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Question Number 116226 by bobhans last updated on 02/Oct/20
(1/(2+(√2))) +(1/(3(√2)+2(√3))) +(1/(4(√3)+3(√4)))+...+(1/(100(√(99))+99(√(100))))
$$\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}}\:+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{4}}}+…+\frac{\mathrm{1}}{\mathrm{100}\sqrt{\mathrm{99}}+\mathrm{99}\sqrt{\mathrm{100}}} \\ $$
Commented by bemath last updated on 02/Oct/20
0.9
$$\mathrm{0}.\mathrm{9} \\ $$
Answered by john santu last updated on 02/Oct/20
(1/(2+(√2)))×((2−(√2))/(2−(√2))) = ((2−(√2))/2)=1−((√2)/2) (1/(3(√2)+2(√3) ))×((3(√2)−2(√3))/(3(√2)−2(√3))) = ((3(√2)−2(√3))/6)=((√2)/2)−((√3)/3) (1/(4(√3)+3(√4)))×((4(√3)−3(√4))/(4(√3)−3(√4))) = ((4(√3)−3(√4))/(12))=((√3)/3)−((√4)/4) then = 1−((√2)/2)+((√2)/2)−((√3)/3)+((√3)/3)−((√4)/4)+...+((√(99))/(99))−((√(100))/(100)) (telescoping series) = 1−((√(100))/(100)) = 1−(1/(10)) = (9/(10)) = 0.9
$$\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{2}}}×\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\:=\:\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}\:}×\frac{\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{6}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{4}}}×\frac{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{4}}}{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{4}}}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{4}}}{\mathrm{12}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}−\frac{\sqrt{\mathrm{4}}}{\mathrm{4}} \\ $$$${then}\: \\ $$$$=\:\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}−\frac{\sqrt{\mathrm{4}}}{\mathrm{4}}+…+\frac{\sqrt{\mathrm{99}}}{\mathrm{99}}−\frac{\sqrt{\mathrm{100}}}{\mathrm{100}} \\ $$$$\left({telescoping}\:{series}\right) \\ $$$$=\:\mathrm{1}−\frac{\sqrt{\mathrm{100}}}{\mathrm{100}}\:=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}\:=\:\frac{\mathrm{9}}{\mathrm{10}}\:=\:\mathrm{0}.\mathrm{9} \\ $$
Answered by Dwaipayan Shikari last updated on 02/Oct/20
(1/(2+(√2)))=((2−(√2))/2)=1−(1/( (√2))) (1/(3(√2)+2(√3)))=((3(√2)−2(√3))/6)=(1/( (√2)))−(1/( (√3))) So (1/(2+(√2)))+(1/(3(√2)+2(√3)))+...=1−(1/( (√2)))+(1/( (√2)))−(1/( (√3)))+.....−(1/( (√(100)))) =1−(1/(10))=(9/(10))
$$\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{2}}}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}}=\frac{\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{6}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{So} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}}+…=\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+…..−\frac{\mathrm{1}}{\:\sqrt{\mathrm{100}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}=\frac{\mathrm{9}}{\mathrm{10}} \\ $$

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