方程与函数21.已知f(3^x)=4x[(log2)3]+233,求f(2)+f(4)+f(8)+...+f(2^8)的值.

Question

方程与函数2
1.已知f(3^x)=4x[(log2)3]+233,求f(2)+f(4)+f(8)+...+ f(2^8) 的值.

Answer 1

令 t = 3^x ,得 x = (log3)t = lgt / lg3 ,
所以 f(t) = f(3^x) = 4 [(log3)t] [(log2)3] + 233
= 4 (lgt / lg3) (lg3 / lg2) + 233
= 4 lg t / lg 2 + 233
所以 f(2^n) = 4 lg (2^n) / lg 2 + 233
= 4 n + 233 ,
所以 f(2) + f(4) + f(8) + ...+ f(2^8)
= 4 * 1 + 233 + 4 * 2 + 233 + 4 * 3 + 233 + … + 4 * 8 + 233
= 4 * (1 + 2 + 3 + … + 8) + 233 * 8
= 4 * (1 + 8) * 8 / 2 + 233 * 8
= 8 * (18 + 233)
= 8 * 251
= 2008