Because this is a multiple-choice question, it can be done by the undetermined coefficient method:

Let n=3, and from the point of view of the topic, you can go up one or two levels at a time, so f(3)=3, f(2)=2, f( 1)= 1.

In this way, a or d is right. At this time, the f(4) test shows that f(4)=5, so D is right.

The second question:

An+ 1=an/ 1+2an。 This formula is complicated. You can find 1/an+ 1 first.

So 1/an+1 =1+2an/an =1/an+1,so the sequence {1/an} is a head item1.

The third question:

I first thought of this question as c 1=2, and I haven't thought about it since. So go back to the original method.

C3=a 1+2d+b 1q2 (series {a} is arithmetic and d is tolerance; {b} is proportional, q is common ratio, and q is followed by exponent)

C4=a 1+3d+b 1q3 is solved simultaneously, and d= 1 q=2 is obtained (these two equations are difficult to solve, so it is recommended to solve them by observation, and the equal sign holds when d= 1 q=2). I won't write it later, but summarize it in groups.

The fourth question:

I don't want to write. Tell me a train of thought. The title requires {an-3an- 1} to be a geometric series.

So if an-3an-1= (3n-1)-6an-1increases the common factor by -5, it can be done.

Later, it was done by reverse recursion.