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