Question Number 4941 by madscientist last updated on 25/Mar/16
$${if}\:{i}\:{had}\:{a}\:{grid}\:{with}\:\mathrm{4}\:{quadrants}\:{and}\:{i}\:{shaded}\:{a}\:\: \\ $$$${quarter}\:{of}\:{it}\:{and}\:{chose}\:{one}\:{of}\:{the}\:\mathrm{4}\:{quadrants}\:{to}\:{form}\:{a}\:{new}\:{grid} \\ $$$${and}\:{shaded}\:{it}\:{and}\:{repeated}\:{the}\:{process}\:{infinitly}\:{what}\: \\ $$$${is}\:{the}\:{amount}\:{of}\:{the}\:{shaded}\:{area}? \\ $$$${and}\:{what}\:{formula}\:{will}\:{solve}\:{it}? \\ $$
Answered by FilupSmith last updated on 25/Mar/16
$${At}\:{first}:\:\frac{\mathrm{1}}{\mathrm{4}}\:{shaded} \\ $$$${then}\:{another}\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$… \\ $$$$ \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{i}} },\:\:\:\:\:\:\:{ratio}\:=\:{r}\:=\:\frac{\mathrm{1}}{\mathrm{4}},\:\:\mathrm{1}{st}\:{term}\:=\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\therefore\:{S}\:=\:\frac{{a}}{\mathrm{1}−{r}}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\frac{\mathrm{3}}{\mathrm{4}}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${S}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$