Question

$\frac{8}{2^2+1}+\frac{16}{2^6+1}+\frac{32}{2^{14}+1}+\frac{64}{2^{14}+1}+\frac{128}{2^{18}+1}=$ ?

Answer 1

We have,

$\begin{array}{c}\frac{8}{2^2+1}+\frac{16}{2^6+1}+\frac{32}{2^{14}+1}+\frac{64}{2^{14}+1}+\frac{128}{2^{18}+1}\\ \frac{32}{2^{14}+1}+\frac{64}{2^{14}+1}:\frac{96}{16385}\\ \frac{8}{2^2+1}+\frac{16}{2^6+1}+\frac{96}{16385}+\frac{128}{2^{18}+1}\\ 2^2+1=5\\ =\frac{8}{5}+\frac{16}{2^6+1}+\frac{96}{16385}+\frac{128}{2^{18}+1}\\ 2^6+1=65\\ =\frac{8}{5}+\frac{16}{65}+\frac{96}{16385}+\frac{128}{2^{18}+1}\\ Leastcommonmultiplierof5,65,16385:859049165\\ =\frac{1374478664}{859049165}+\frac{211458256}{859049165}+\frac{5033184}{859049165}+\frac{419456}{859049165}\\ Sincethedenominatorareequal,combinethefraction:\\ \frac{a}{b}\ne \frac{b}{c}=\frac{a\ne b}{c}\\ =\frac{1374478664+211458256+5033184}{859049165}\\ =\frac{1591389560}{859049165}\\ =\frac{318277912}{171809833}\end{array}$