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Question-67826




Question Number 67826 by peter frank last updated on 31/Aug/19
Commented by gunawan last updated on 01/Sep/19
x=1+a+a^2 +a^3 +..  x=(1/(1−a)) ⇒1− a=(1/x)⇒ a=1−(1/x)=((x−1)/x)  y=1+b+b^2 +b^3 +...  y=(1/(1−b)) ⇒ 1−b=(1/y) ⇒b=((y−1)/y)   1+(ab)+(ab)^2 +(ab)^3 +..  =(1/(1−ab))=(1/(((xy)/(xy))−((((x−1)(y−1))/(xy)))))  =(1/((((xy−(xy−x−y+1))/(xy)))))=((xy)/(x+y−1))
$${x}=\mathrm{1}+{a}+{a}^{\mathrm{2}} +{a}^{\mathrm{3}} +.. \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{1}−{a}}\:\Rightarrow\mathrm{1}−\:{a}=\frac{\mathrm{1}}{{x}}\Rightarrow\:{a}=\mathrm{1}−\frac{\mathrm{1}}{{x}}=\frac{{x}−\mathrm{1}}{{x}} \\ $$$${y}=\mathrm{1}+{b}+{b}^{\mathrm{2}} +{b}^{\mathrm{3}} +… \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{1}−{b}}\:\Rightarrow\:\mathrm{1}−{b}=\frac{\mathrm{1}}{{y}}\:\Rightarrow{b}=\frac{{y}−\mathrm{1}}{{y}}\: \\ $$$$\mathrm{1}+\left({ab}\right)+\left({ab}\right)^{\mathrm{2}} +\left({ab}\right)^{\mathrm{3}} +.. \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{ab}}=\frac{\mathrm{1}}{\frac{{xy}}{{xy}}−\left(\frac{\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)}{{xy}}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(\frac{{xy}−\left({xy}−{x}−{y}+\mathrm{1}\right)}{{xy}}\right)}=\frac{{xy}}{{x}+{y}−\mathrm{1}} \\ $$
Commented by mathmax by abdo last updated on 01/Sep/19
its not correct if ∣a∣≥1 or  ∣b∣≥1...!
$${its}\:{not}\:{correct}\:{if}\:\mid{a}\mid\geqslant\mathrm{1}\:{or}\:\:\mid{b}\mid\geqslant\mathrm{1}…! \\ $$
Commented by peter frank last updated on 01/Sep/19
thank you
$${thank}\:{you}\: \\ $$
Commented by peter frank last updated on 01/Sep/19
thank you
$${thank}\:{you}\: \\ $$
Answered by mind is power last updated on 01/Sep/19
x=(1/(1−a))⇒a=((x−1)/x)  y=(1/(1−b))⇒b=((y−1)/y)  ⇒1+ab+a^2 b^2 +.....=Σ(ab)^k =((((y−1)/y))(((x−1)/x)))^k =(1/(1−(((x−1)/x))(((y−1)/y))))=((xy)/(xy−xy+y+x−1))=((xy)/(x+y−1))
$${x}=\frac{\mathrm{1}}{\mathrm{1}−{a}}\Rightarrow{a}=\frac{{x}−\mathrm{1}}{{x}} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{1}−{b}}\Rightarrow{b}=\frac{{y}−\mathrm{1}}{{y}} \\ $$$$\Rightarrow\mathrm{1}+{ab}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} +…..=\Sigma\left({ab}\right)^{{k}} =\left(\left(\frac{{y}−\mathrm{1}}{{y}}\right)\left(\frac{{x}−\mathrm{1}}{{x}}\right)\right)^{{k}} =\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{{x}−\mathrm{1}}{{x}}\right)\left(\frac{{y}−\mathrm{1}}{{y}}\right)}=\frac{{xy}}{{xy}−{xy}+{y}+{x}−\mathrm{1}}=\frac{{xy}}{{x}+{y}−\mathrm{1}} \\ $$
Answered by rrebo5637@gmail.com last updated on 01/Sep/19

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