Let f(x)=2^x+2+162^2x+1+2^x+4+32.

Question

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

Accepted Answer

$\begin{gathered} f(x)=\frac{{42}^{x}+16}{{2.2}^{2x}+{16.2}^{x}+32} \ f(x)=\frac{2({2}^{x}+4)}{{2}^{2x}+{8.2}^{x}+16} \ f(x)=\frac{2}{{2}^{x}+4} \ f(4-x)=\frac{{2}^{x}}{2({2}^{x}+4)} \ f(x)+f(4-x)=\frac{1}{2} \end{gathered}$

So, $\, f(\frac{1}{15})+f(\frac{59}{15})=\frac{1}{2}$

$\begin{gathered} =f(\frac{29}{15})+f(\frac{31}{15})=\frac{1}{2} \ f(\frac{30}{15})=f(2)=\frac{2}{{2}^{2}+4}=\frac{2}{8}=\frac{1}{4} \ \Rightarrow 8(29	imes \frac{1}{2}+\frac{1}{4}) \end{gathered}$

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