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Question Number 105306 by bobhans last updated on 27/Jul/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 28/Jul/20
(1/(2(√1)+1(√2)))+(1/(3(√2)+2(√3)))+(1/(4(√3)+3(√4)))+...+(1/(100(√(99))+99(√(100)))) a_n = (1/((n+1)(√n)+n(√(n+1)))) = (1/( (√(n+1))((√(n^2 +n))+n))) = (1/( (√((n+1)^2 n))+(√(n^2 (n+1))))) = (1/( (√(n.(n+1))))) .[(1/( (√(n+1))+(√n)))] = (((√(n+1))−(√n))/( (√n).(√(n+1)))) = (1/( (√n))) − (1/( (√(n+1)))) telescoping series Σ_(n = 1) ^(99) [ (1/( (√n))) − (1/( (√(n+1)))) ] = (1/( (√1))) − (1/( (√(100)))) = 1−(1/(10)) = 0.9 {★}
$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}}+\mathrm{1}\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{4}}}+…+\frac{\mathrm{1}}{\mathrm{100}\sqrt{\mathrm{99}}+\mathrm{99}\sqrt{\mathrm{100}}} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\sqrt{{n}}+{n}\sqrt{{n}+\mathrm{1}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}\left(\sqrt{{n}^{\mathrm{2}} +{n}}+{n}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\left({n}+\mathrm{1}\right)^{\mathrm{2}} {n}}+\sqrt{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)}}\: \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{{n}.\left({n}+\mathrm{1}\right)}}\:.\left[\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}+\sqrt{{n}}}\right] \\ $$$$=\:\frac{\sqrt{{n}+\mathrm{1}}−\sqrt{{n}}}{\:\sqrt{{n}}.\sqrt{{n}+\mathrm{1}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\:−\:\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}} \\ $$$${telescoping}\:{series}\: \\ $$$$\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\left[\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\:−\:\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}}\:\right]\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}}}\:−\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{100}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}\:=\:\mathrm{0}.\mathrm{9} \\ $$$$\left\{\bigstar\right\} \\ $$
Answered by Dwaipayan Shikari last updated on 27/Jul/20
T_n =(1/((n+1)(√n)+n(√(n+1))))=(1/( (√(n^2 +n)))).(1/(((√(n+1))+(√n))))(By observation) T_n =(((√(n+1))−(√n))/( (√(n^2 +n))))=(1/( (√n)))−(1/( (√(n+1)))) ΣT_(99) =(1/1)−(1/( (√2)))+(1/( (√2)))+...−(1/( (√(99+1)))) ΣT_(99) =1−(1/(10))=(9/(10)) Another way T_1 =(1/(2+(√2)))=((2−(√2))/(4−2))=1−(1/( (√2))) T_2 =(1/(3(√2)+2(√3)))=(1/( (√6)))((1/( (√3)+(√2))))=(1/( (√6)))((√3)−(√2))=(1/( (√2)))−(1/( (√3))) .... T_(99) =(1/(100(√(99))+99(√(100))))=(1/( (√(9900))))((1/( (√(100))))+(1/( (√(99)))))=(1/( (√(9900))))((√(100))−(√(99))) =(1/( (√(99))))−(1/( (√(100)))) Σ^(99) T_n =T_1 +T_2 +....+T_(99) =1−(1/( (√2)))+(1/( (√2)))−(1/( (√3)))+....+(1/( (√(98))))−(1/( (√(99))))+(1/( (√(99))))−(1/( (√(100)))) =1−(1/( (√(100))))=(9/(10))
$${T}_{{n}} =\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\sqrt{{n}}+{n}\sqrt{{n}+\mathrm{1}}}=\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +{n}}}.\frac{\mathrm{1}}{\left(\sqrt{{n}+\mathrm{1}}+\sqrt{{n}}\right)}\left({By}\:{observation}\right) \\ $$$${T}_{{n}} =\frac{\sqrt{{n}+\mathrm{1}}−\sqrt{{n}}}{\:\sqrt{{n}^{\mathrm{2}} +{n}}}=\frac{\mathrm{1}}{\:\sqrt{{n}}}−\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}} \\ $$$$\Sigma{T}_{\mathrm{99}} =\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+…−\frac{\mathrm{1}}{\:\sqrt{\mathrm{99}+\mathrm{1}}} \\ $$$$\Sigma{T}_{\mathrm{99}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}=\frac{\mathrm{9}}{\mathrm{10}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{Another}}\:\boldsymbol{\mathrm{way}} \\ $$$$\boldsymbol{\mathrm{T}}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{2}}}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{4}−\mathrm{2}}=\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\boldsymbol{\mathrm{T}}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$…. \\ $$$$\boldsymbol{\mathrm{T}}_{\mathrm{99}} =\frac{\mathrm{1}}{\mathrm{100}\sqrt{\mathrm{99}}+\mathrm{99}\sqrt{\mathrm{100}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{9900}}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{100}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{99}}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{9900}}}\left(\sqrt{\mathrm{100}}−\sqrt{\mathrm{99}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{99}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{100}}} \\ $$$$\overset{\mathrm{99}} {\sum}\boldsymbol{\mathrm{T}}_{\boldsymbol{\mathrm{n}}} =\boldsymbol{\mathrm{T}}_{\mathrm{1}} +\boldsymbol{\mathrm{T}}_{\mathrm{2}} +….+\boldsymbol{\mathrm{T}}_{\mathrm{99}} =\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+….+\frac{\mathrm{1}}{\:\sqrt{\mathrm{98}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{99}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{99}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{100}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{100}}}=\frac{\mathrm{9}}{\mathrm{10}} \\ $$

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