Rational beauty!!

$\frac{225}{157}=[a;b,c,d,e]\;means\;a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}=\frac{225}{157}$ So we have to convert $\frac{225}{157}$ to continued fraction and then find out the values of $a,b,c,d\;and\;e$ So let,s start: $\frac{225}{157}=1\frac{68}{157}=1+\frac{68}{157}=1+\frac{1}{\frac{157}{68}}$ $\frac{157}{68}=2\frac{21}{68}=2+\frac{21}{68}=2+\frac{1}{\frac{68}{21}}$ $\frac{68}{21}=3\frac{5}{21}=3+\frac{5}{21}=3+\frac{1}{\frac{21}{5}}$ $\frac{21}{5}=4\frac{1}{5}=4+\frac{1}{5}\;(Here\;process\;stops)$ Observe that each of these fractions can be substituted into the previous fraction.Combining all of these ,we get: $\frac{225}{157}=1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5}}}}$ Equating this with $a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}$ , We get $a=1,b=2,c=3,d=4\;and\;e=5$ Adding all these,we get $a+b+c+d+e=1+2+3+4+5=\frac{5\times6}{2}=\frac{30}{2}=\boxed{15}$ If anyone wants to know more about continued fractions,he can do so here