China Mathematical Competition

Solution I We have $$\begin{array}{rl}f(x+2)-f(x)& =-(f(x+4)-f(x+2))-(f(x+6)-f(x+4))+(f(x+6)-f(x))\\ & \ge -3\times 2^{x+2}-3\times 2^{x+4}+63\times 2^x=3\times 2^x.\end{array}$$ This means that $f(x+2)-f(x)=3\times 2^x$. So we have $$\begin{array}{rl}f(2008)& =f(2008)-f(2006)+f(2006)-f(2004)+\dots \\ & +f(2)-f(0)+f(0)\\ & =3\times (2^{2006}+2^{2004}+\cdots +2^2+1)+f(0)\\ & =3\times \frac{4^{1003}+1}{4-1}+2008\\ & =2^{2008}+2007.\end{array}$$

Solution II We define $g(x)=f(x)-2^x$. Then we have $$\begin{array}{rl}g(x+2)-g(x)& =f(x+2)-f(x)-2^{x+2}+2^x\\ & \le 3\times 2^x-3\times 2^x=0,\\ g(x+6)-g(x)& =f(x+6)-f(x)-2^{x+6}+2^x\\ & \ge 63\times 2^x-63\times 2^x=0.\end{array}$$ This means that $g(x)\le g(x+6)\le g(x+4)\le g(x+2)\le g(x)$, and it implies that $g(x)$ is a periodic function with 2 as a period. So $$\begin{array}{rl}f(2008)& =g(2008)+2^{2008}=g(0)+2^{2008}\\ & =2007+2^{2008}.\end{array}$$