Minhang District Mathematical Model Volume 2 20 16

Solution: Because f (x+2)-f (x) ≤ 3 x ①,

So f (x+4)-f (x+2) ≤ 3 (x+2) = 3x * 32 = 9 * 3x②.

F (x+4)-F (x) ≤ 10 * 3 x ③ from ①+②.

And because f (x+4)-f (x) ≥ 10 * 3 x, ④.

Comparing ③ with ④, we get f (x+4)-f (x) = 10 * 3 x.

f(20 14)=f(20 10)+ 10*3^20 10=f(2006)+ 10*(3^2006+3^20 10)=……

= f (2)+10 * (3 2+3 6+310+) ...+3 2006+3 2010) (the first item in brackets is 9, and the public ratio is 3 4 503.

=f(0)+3^0+ 10*[9*(3^4)^503- 1]/(3^4- 1)= 1/8+ 1+[3^20 14-9]/8=3^20 14/8

Is it okay? Hope to adopt! ! !