Let $f(x)=\frac{2^{x+2}+16}{2^{2 x+1}+2^{x+4}+32}$. Then the value of $8\left(f\left(\frac{1}{15}\ri

Question

Let $f(x)=\frac{2^{x+2}+16}{2^{2 x+1}+2^{x+4}+32}$. Then the value of $8\left(f\left(\frac{1}{15}ight)+f\left(\frac{2}{15}ight)+\ldots+f\left(\frac{59}{15}ight)ight)$ is equal to\

Professor PC Thomas Classes · Answer

(a) : $f(x)=\frac{2^{x+2}+16}{2^{2 x+1}+2^{x+4}+32}$\ $=\frac{2^{2}\left(2^{x}+4ight)}{2\left[\left(2^{x}ight)^{2}+8 	imes 2^{x}+16ight)}=\frac{2\left(2^{x}+4ight)}{\left(2^{x}+4ight)^{2}}$ $$ \begin{aligned} & =\frac{2}{2^{x}+4} \ & \Rightarrow f(4-x)=\frac{2}{2^{4-x}+4}=\frac{2^{x}}{2\left(2^{x}+4ight)} \ & f(x)+f(4-x)=\frac{2}{2^{x}+4}+\frac{2^{x}}{2\left(2^{x}+4ight)}=\frac{1}{2} \ & 	herefore f\left(\frac{1}{15}ight)+f\left(\frac{59}{15}ight)=f\left(\frac{2}{15}ight)+f\left(\frac{58}{15}ight) \ & =f\left(\frac{3}{15}ight)+f\left(\frac{57}{15}ight)=\ldots=f\left(\frac{29}{15}ight)+f\left(\frac{31}{15}ight)=\frac{1}{2} \ & f\left(\frac{30}{15}ight)=f(2)=\frac{2}{4+4}=\frac{1}{4} \ & 8\left[f\left(\frac{1}{15}ight)+f\left(\frac{2}{15}ight)+\ldots .+f\left(\frac{59}{15}ight)ight] \ & =8\left[\frac{1}{2}+\frac{1}{2}+\ldots . . 	ext { upto (29 terms) }+\frac{1}{4}ight] \ & =8\left(\frac{29}{2}+\frac{1}{4}ight)=118 \end{aligned} $$