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Question Number 45283 by peter frank last updated on 11/Oct/18
if x=1+a+a^2 +a^3 +………      y=1+b+b^2 +b^3 +………  prove that  1+ab+a^2 b^2 +a^3 b^3 +………=((xy)/(x+y−1))
$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{x}}=\mathrm{1}+\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\ldots\ldots\ldots \\ $$$$\:\:\:\:\boldsymbol{\mathrm{y}}=\mathrm{1}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\ldots\ldots\ldots \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$$$\mathrm{1}+\boldsymbol{\mathrm{ab}}+\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} \boldsymbol{\mathrm{b}}^{\mathrm{3}} +\ldots\ldots\ldots=\frac{\boldsymbol{\mathrm{xy}}}{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}−\mathrm{1}} \\ $$
Answered by ajfour last updated on 11/Oct/18
x=(1/(1−a))  ;  y= (1/(1−b))  z= 1+ab+a^2 b^2 +a^3 b^3 +...    = (1/(1−ab)) = (1/(1−(1−(1/x))(1−(1/y))))    = ((xy)/(xy−(x−1)(y−1)))    = ((xy)/(x+y−1)) .
$${x}=\frac{\mathrm{1}}{\mathrm{1}−{a}}\:\:;\:\:{y}=\:\frac{\mathrm{1}}{\mathrm{1}−{b}} \\ $$$${z}=\:\mathrm{1}+{ab}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{3}} {b}^{\mathrm{3}} +… \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{1}−{ab}}\:=\:\frac{\mathrm{1}}{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{y}}\right)} \\ $$$$\:\:=\:\frac{{xy}}{{xy}−\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)} \\ $$$$\:\:=\:\frac{{xy}}{{x}+{y}−\mathrm{1}}\:. \\ $$
Commented by peter frank last updated on 11/Oct/18
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by maxmathsup by imad last updated on 11/Oct/18
this is true if we know that ∣a∣<1 and ∣b∣<1...
$${this}\:{is}\:{true}\:{if}\:{we}\:{know}\:{that}\:\mid{a}\mid<\mathrm{1}\:{and}\:\mid{b}\mid<\mathrm{1}… \\ $$

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