Second year: B2 = [b1-x] (1+0.25) = a (1+0.25) 2-x (1+0.25).

Third year: B3 = [B2-x] (1+0.25) = a (1+0.25) 3-x (1+0.25) 2-x (1+0.25).

You can get it by analogy.

bn=a( 1+0.25)^n-x( 1-( 1+0.25)^(n-2))/( 1-( 1+0.25))

Well, the key points have been worked out. (1)(2)(3) Just put the data in.

Ok, by the way, help you solve the following problems, and give me more if you are happy ~

A. because An=log _2(Bn), bn = 2 an.

Because An is arithmetic progression, Bn:B(n- 1) = 2 [an-a (n-1)] = 2d (D is the arithmetic of an * * (I can't remember what it's called)), so BN: b (n-1).

Sn=a 1n+n(n- 1)d/2

Given S 10 and a 1, if d is calculated, the general term 2 D of Bn can be calculated, and an = log _ 2 (Bn). Because A 1 is known, the general term formula of Bn can be calculated. The sum of the first five terms of Bn can be obtained by the summation formula of geometric series.

B. 1。 Just make sure that (-x 2+2x+3) is greater than 0, so that X is within-1~3.

2. The monotonicity of the composite function is sufficient, and the log- 1/2X monotonically decreases, so as long as the monotonically decreasing interval of (-x 2+2x+3) is found, the monotonically decreasing interval of the total function is (-x 2+2x+3), which is calculated as: 1~3.

C. use the first n terms of geometric series and the formula to solve.

Sn=an。 ( 1-q^(n- 1))/( 1-q)

Go down and find q for A 1 and S3.

The others are ok.