:-s

N

noinhobinhyen

$A=0,050505... = \dfrac{5}{100}+\dfrac{5}{100^2}+\dfrac{5}{100^3}+...+
\dfrac{5}{100^n}$

$A=\dfrac{5}{100}.\dfrac{1-\dfrac{1}{100^n}}{1-\dfrac{1}{100}}$

$LimA= \dfrac{5}{100}.\dfrac{1}{\dfrac{99}{100}} = \dfrac{5}{99}$
 
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