$Если i=0.2$ $$P=\frac{100}{1.2}+\frac{100}{1.2^2}+\frac{100}{1.2^3}+\frac{100}{1.2^4}+\frac{100}{1.2^5}$$ $$P=\frac{100}{1-\frac{1}{1.2}}=\frac{100}{\frac{0.2}{1.1}}=550$$
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$$P=\frac{100}{(1+0.1)}+\frac{100}{(1+0.1)^2}+\frac{100}{(1+0.1)^3}+\frac{100}{(1+0.1)^4}+\frac{100}{(1+0.1)^5}$$
$$P=\frac{100}{1.1}+\frac{100}{1.1^2}+\frac{100}{1.1^3}+\frac{100}{1.1^4}+\frac{100}{1.1^5}$$
$$P=\frac{100}{1-\frac{1}{1.1}}=\frac{100}{\frac{0.1}{1.1}}=1100$$
$Если i=0.2$
$$P=\frac{100}{1.2}+\frac{100}{1.2^2}+\frac{100}{1.2^3}+\frac{100}{1.2^4}+\frac{100}{1.2^5}$$
$$P=\frac{100}{1-\frac{1}{1.2}}=\frac{100}{\frac{0.2}{1.1}}=550$$