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Question-202104




Question Number 202104 by necx122 last updated on 20/Dec/23
Commented by necx122 last updated on 20/Dec/23
please help with this
$${please}\:{help}\:{with}\:{this} \\ $$
Commented by cortano12 last updated on 21/Dec/23
  S =Σ_(i=1) ^(99) ((1/( (√i)+(√(i+1)))))= Σ_(i=1) ^(99) ((((√i)−(√(i+1)))/(−1)))   ( telescopic series )    S = −((√1)−(√(100))) =−(−9)=9
$$\:\:\mathrm{S}\:=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{i}}+\sqrt{\mathrm{i}+\mathrm{1}}}\right)=\:\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\left(\frac{\sqrt{\mathrm{i}}−\sqrt{\mathrm{i}+\mathrm{1}}}{−\mathrm{1}}\right) \\ $$$$\:\left(\:\mathrm{telescopic}\:\mathrm{series}\:\right) \\ $$$$\:\:\mathrm{S}\:=\:−\left(\sqrt{\mathrm{1}}−\sqrt{\mathrm{100}}\right)\:=−\left(−\mathrm{9}\right)=\mathrm{9}\: \\ $$
Answered by esmaeil last updated on 20/Dec/23
(((√1)−(√2))/(−1))+(((√2)−(√3))/(−1))+(((√3)−(√4))/(−1))+...+(((√(99))−(√(100)))/(−1))=  =9
$$\frac{\sqrt{\mathrm{1}}−\sqrt{\mathrm{2}}}{−\mathrm{1}}+\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}}{−\mathrm{1}}+\frac{\sqrt{\mathrm{3}}−\sqrt{\mathrm{4}}}{−\mathrm{1}}+…+\frac{\sqrt{\mathrm{99}}−\sqrt{\mathrm{100}}}{−\mathrm{1}}= \\ $$$$=\mathrm{9} \\ $$
Commented by necx122 last updated on 20/Dec/23
Sincerely, I still do not understand. At forst  I thought the conjugate of each fraction  was taken and that′s how it was done  for all but not at all. Please shed some  more light anyone
$${Sincerely},\:{I}\:{still}\:{do}\:{not}\:{understand}.\:{At}\:{forst} \\ $$$${I}\:{thought}\:{the}\:{conjugate}\:{of}\:{each}\:{fraction} \\ $$$${was}\:{taken}\:{and}\:{that}'{s}\:{how}\:{it}\:{was}\:{done} \\ $$$${for}\:{all}\:{but}\:{not}\:{at}\:{all}.\:{Please}\:{shed}\:{some} \\ $$$${more}\:{light}\:{anyone} \\ $$
Commented by esmaeil last updated on 20/Dec/23
(1/( (√1)+(√2)))+(1/( (√2)+(√3)))+...+(1/( (√(99))+(√(100))))=A  (1/( (√a)+(√b)))=(1/( (√a)+(√b)))×(((√a)−(√b))/( (√a)−(√b)))=(((√a)−(√b))/(a−b)) ★  (1/( (√1)+(√2)))=^★ (((√1)−(√2))/(1−2=−1)),  (1/( (√2)+(√3)))=^★ (((√2)−(√3))/(2−3=−1)),  ...,(1/( (√(99))+(√(100))))=^★ (((√(99))−(√(100)))/(99−100=−1))  ⇒  A=(((√1)−(√2)+(√2)−(√3)+(√3)−(√4)...−(√(100)))/(−1))  =((1−(√(100)))/(−1))=9
$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}}+\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}}+…+\frac{\mathrm{1}}{\:\sqrt{\mathrm{99}}+\sqrt{\mathrm{100}}}={A} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{a}}+\sqrt{{b}}}=\frac{\mathrm{1}}{\:\sqrt{{a}}+\sqrt{{b}}}×\frac{\sqrt{{a}}−\sqrt{{b}}}{\:\sqrt{{a}}−\sqrt{{b}}}=\frac{\sqrt{{a}}−\sqrt{{b}}}{{a}−{b}}\:\bigstar \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}}+\sqrt{\mathrm{2}}}\overset{\bigstar} {=}\frac{\sqrt{\mathrm{1}}−\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{2}=−\mathrm{1}},\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}}\overset{\bigstar} {=}\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}}{\mathrm{2}−\mathrm{3}=−\mathrm{1}}, \\ $$$$…,\frac{\mathrm{1}}{\:\sqrt{\mathrm{99}}+\sqrt{\mathrm{100}}}\overset{\bigstar} {=}\frac{\sqrt{\mathrm{99}}−\sqrt{\mathrm{100}}}{\mathrm{99}−\mathrm{100}=−\mathrm{1}} \\ $$$$\Rightarrow \\ $$$${A}=\frac{\sqrt{\mathrm{1}}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{4}}…−\sqrt{\mathrm{100}}}{−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}−\sqrt{\mathrm{100}}}{−\mathrm{1}}=\mathrm{9} \\ $$
Commented by necx122 last updated on 21/Dec/23
Thanks so much.
Answered by MM42 last updated on 20/Dec/23
s=((√2)−(√1))+((√3)−(√2))+((√4)−(√3))+...+((√(99))−(√(98)))+((√(100))−(√(99)))  =(√(100))−(√1)= 9 ✓
$${s}=\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{1}}\right)+\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)+\left(\sqrt{\mathrm{4}}−\sqrt{\mathrm{3}}\right)+…+\left(\sqrt{\mathrm{99}}−\sqrt{\mathrm{98}}\right)+\left(\sqrt{\mathrm{100}}−\sqrt{\mathrm{99}}\right) \\ $$$$=\sqrt{\mathrm{100}}−\sqrt{\mathrm{1}}=\:\mathrm{9}\:\checkmark \\ $$
Commented by MathematicalUser2357 last updated on 22/Dec/23
How did you get ✓
$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\checkmark \\ $$

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