Question 27: The function $f(x)$ with domain $R$ satisfies: when $x \in[0,1)$, $f(x)=2^{x}-x$, and for any real number $x$, $f(x)+f(x+1)=1$. Let $a=\log _{2} 3$, then the value of the expression $f(a)+f(2 a)+f(3 a)$ is $\qquad$ -
Question 27, Answer: The answer is $\frac{17}{16}^{\circ}$.
Notice that $\mathrm{a}=\log _{2} 3 \in[1,2), 2 a=\log _{2} 9 \in[3,4), 3 a=\log _{2} 27 \in[4,5)$. Therefore, we know $f(a)=1-f(a-1)=1-\left(2^{a-1}-(a-1)\right)=-\frac{3}{2}+\log _{2} 3$
$f(2 a)=1-f(2 a-1)=f(2 a-2)=1-f(2 a-3)=1-\left(2^{2 a-3}-(2 a-3)\right)=-\frac{25}{8}+$ $2 \log _{2} 3$
$$
f(3 a)=1-f(3 a-1)=f(3 a-2)=1-f(3 a-3)=f(3 a-4)=\left(2^{3 a-4}-(3 a-4)\right)=
$$
$\frac{91}{16}-3 \log _{2} 3$
Thus, $\mathrm{f}(\mathrm{a})+\mathrm{f}(2 \mathrm{a})+\mathrm{f}(3 \mathrm{a})=\frac{91}{16}-\frac{3}{2}-\frac{25}{8}=\frac{17}{16}$.