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Question-174361




Question Number 174361 by infinityaction last updated on 30/Jul/22
Commented by a.lgnaoui last updated on 31/Jul/22
S=((5×3)/((5−3)(5^2 −3^2 )))+((5^2 ×3^2 )/((5^2 −3^2 )(5^3 −3^3 )))+((5^3 ×3^3 )/((5^3 −3^3 )(5^4 −3^4 )))+....+((5^n ×3^n )/((5^n −3^n )(5^(n+1) −3^(n+1) )))  =Σ_(n=1) ^∞ ((5^n 3^n )/((5^n −3^n )(5^(n+1) −3^(n+1) )))  ((5^n 3^n )/((5^n −3^n )(5^(n+1) −3^(n+1) )))=(1/(((1/3^n )−(1/5^n ))((1/3^(n+1) )−(1/5^(n+1) ))))  S=Σ_(n=1) ^∞ [(1/(((1/3^n )−(1/5^n ))((1/3^(n+1) )−(1/5^(n+1) ))))]  ........
$${S}=\frac{\mathrm{5}×\mathrm{3}}{\left(\mathrm{5}−\mathrm{3}\right)\left(\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)}+\frac{\mathrm{5}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} }{\left(\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{5}^{\mathrm{3}} −\mathrm{3}^{\mathrm{3}} \right)}+\frac{\mathrm{5}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{3}} }{\left(\mathrm{5}^{\mathrm{3}} −\mathrm{3}^{\mathrm{3}} \right)\left(\mathrm{5}^{\mathrm{4}} −\mathrm{3}^{\mathrm{4}} \right)}+….+\frac{\mathrm{5}^{{n}} ×\mathrm{3}^{{n}} }{\left(\mathrm{5}^{{n}} −\mathrm{3}^{{n}} \right)\left(\mathrm{5}^{{n}+\mathrm{1}} −\mathrm{3}^{{n}+\mathrm{1}} \right)} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{5}^{{n}} \mathrm{3}^{{n}} }{\left(\mathrm{5}^{{n}} −\mathrm{3}^{{n}} \right)\left(\mathrm{5}^{{n}+\mathrm{1}} −\mathrm{3}^{{n}+\mathrm{1}} \right)} \\ $$$$\frac{\mathrm{5}^{{n}} \mathrm{3}^{{n}} }{\left(\mathrm{5}^{{n}} −\mathrm{3}^{{n}} \right)\left(\mathrm{5}^{{n}+\mathrm{1}} −\mathrm{3}^{{n}+\mathrm{1}} \right)}=\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{3}^{{n}} }−\frac{\mathrm{1}}{\mathrm{5}^{{n}} }\right)\left(\frac{\mathrm{1}}{\mathrm{3}^{{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{5}^{{n}+\mathrm{1}} }\right)} \\ $$$${S}=\sum_{{n}=\mathrm{1}} ^{\infty} \left[\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{3}^{{n}} }−\frac{\mathrm{1}}{\mathrm{5}^{{n}} }\right)\left(\frac{\mathrm{1}}{\mathrm{3}^{{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{5}^{{n}+\mathrm{1}} }\right)}\right] \\ $$$$…….. \\ $$

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