Given that the function f (x) = KX + B, f (1) = 2, f (1), f (2), f (4) is a comparative sequence, Let f (n) = log2 (an-2n), find the sequence I'm sorry. (I just found out, I'm sorry.) Find the first n terms and Sn of {an} of sequence

Given that the function f (x) = KX + B, f (1) = 2, f (1), f (2), f (4) is a comparative sequence, Let f (n) = log2 (an-2n), find the sequence I'm sorry. (I just found out, I'm sorry.) Find the first n terms and Sn of {an} of sequence

Because f (1), f (2) and f (4) are in equal proportion sequence, then: F & # 178; (2) = f (1) f (4), that is:
f²(2)=2f(4)
Because f (2) = 2K + B, f (4) = 4K + B, f (1) = K + B = 2
So: (2k + b) & # = 2 (4K + b)
From the above two formulas, we can get: k = 2, B = 0
That is, f (x) = 2x
So f (n) = 2n = log2 (an-2n)
So: an = 4 ^ n + 2n
So:
Sn=4^1+4^2+...+4^n+2(1+2+3+..+n)
=4（1-4^n)/(1-4)+2(1+n)n/2
=4(4^n-1)/3+n(n+1)

In known sequence {an}, A1 = 1, an = an-1.3 ^ (n-1) (n ≥ 2), Let f (n) = log3an / 9 ^ n (n ∈ n *), the sum of the first n terms of sequence {BN} be f (n)
(1) find the general formula of {BN}
(2) find the sum of the first n terms of {BN}

1
f(n)=log3an/9^n
f(n-1)=log3an-1/9^(n-1)
bn=f(n)-f(n-1)=log3an/9^n-log3an-1/9^(n-1)=log3an/9an-1
And: an = an-1.3 ^ (n-1)
bn=log3an/9an-1
＝log3[3^(n-1)/9]=log3[3^(n-3)]=n-3
B1 = f (1) = - 2 also holds,
So BN = n-3
two
b1=-2
S=n*(-2+n-3)/2=n*(n-5)/2

The function f (x) = ln (1 + x) - MX is known. Question (3) let an = 1 / (n + 1) + 1 / (n + 2) +. + 1 / (n + (n + 1)) (n belongs to n *), and prove that an is greater than LN2,

Taking M = 1, we can prove that ln [(n + 1) / N] = ln [1 + (1 / N)] 0
If f '(x) = - X / (1 + x) 0 and f (x) is continuous at x = 0, then f (x)