1.設(shè) f (x) = ax + b； f(1) = a+b = 1
由題意：f (2) + f(5) = 2 x (f (3) + 1)
故 (2a+b) + (5a+b) = 2 x (3a + b + 1),7a + 2b = 6a + 2b + 2,a = 2,b = -1
所以 f(x) = 2x - 1
對(duì)于任何一個(gè) N (正整數(shù)),有 2f(N) = 4 N - 2 = (2(N-1)-1) + (2(N+1)-1) = f(N-1) + f(N+1)
所以an = f(N) 為等差數(shù)列
2.f(x) = 2x-1,f(x+1) = 2x + 1,
中間插入兩個(gè)數(shù)字并保持等差,所以差值d為 2/3,
設(shè)bn = g(n) = 2/3 n + x,由 a(1) = b(1) = 1,有 x = 1/3
所以構(gòu)造成 bn = g(n) = 2/3 n +1/3,
Bn = b1 + b2 + ...+ bn
= (b1 + bn)*n/2 = (1 + 2/3 n + 1/3) * n /2 = 1/3 n^2 + 2/3 n
3.3bn-1 = 3*(2/3n+1/3) = 2n+1
cn = 2^(2n+1)
Cn = c1 + c2 + ...+ cn
= (2^2 - 1) * (cn + ...+ c1) / (2^2 -1)
= c(n+1) - cn + cn - c(n-1) + ...+ c2 - c1)
= (c(n+1) - c1) / 3
= (2^ (2n+3) - 8) / 3