When n= 1, 2, 3, 4, 5, 6 and 7 are brought in, A2 = 3/4, A3 = 2/3, A4 = 5/8, A5 = 3/5, A6 = 7/ 12 and A8 = 4/7.

It can be found that the numerator of odd terms is arithmetic progression whose first term is 1, and the denominator is arithmetic progression whose first term is 1, with a tolerance of 2. Let k be a positive integer, then a (2k-1) = k/[1+(k-1) * 2.

It is also found that the numerator of even terms is arithmetic progression whose first term is 3, and the denominator is arithmetic progression whose first term is 4, with a tolerance of 4. Let k be a positive integer, then a (2k) = [3+(k-1) * 2]/[4+(k-1) * 4] = (1).

Taken together, it is a (n) = (1/2) * (1+1/n).

Of course, this formula is derived from conjecture, and the following will be proved by mathematical induction! (Omitted here, please ask questions if necessary. )

2. Next, find the expression of cn:

bn = 2/(2an- 1)= 2/[2 *( 1/2)*( 1+ 1/n)- 1]= 2/n .

Cn= (radical number 2) bn = 2 (bn/2) = 2 (1/n)

3. The following is proved by reduction to absurdity!

Proof: Obviously, cn is a decreasing sequence (please ask questions if you have any questions). Therefore, if there are three terms ci, cj, ck, ci, cj and ck in the sequence cn, a arithmetic progression can be formed, where I, J and K are positive integers greater than or equal to 1, i.

Then 2cj=ci+ck.

Namely: 2 * 2 (1/j) = 2 (1/i)+2 (1/k), namely: 2 (1+ 1/j) = 2 (.

So: p = p * [2 (1/i-1/j)+2 (1/k-1/j)]-

so 2( 1/I- 1- 1/j)+2( 1/k- 1/j)]= 1。

Let a = 2 (1/I-1-kloc-0//j) and b = 2 (1/k- 1/j)].

If I> 1, then k >；; 1, j> 1, then1/I-1/j; 1/ 1.4 14+ 1/ 1.4 14 & gt； 1.

If i= 1, (this is very complicated, J and K need to be further subdivided, and some cases can be summarized as A+B >; 1, some less than 1, but definitely not equal to 1).

So the contradiction is that 2 (1/I-1-1/j)+2 (1/k-1/j)] =1does not hold.

In addition, this question can be included in the scope of the college entrance examination, and there is nothing out of line. The knowledge involved includes the proof of sequence, monotonicity of function, mathematical induction, the usage of reduction to absurdity and so on. Very comprehensive, but the last question is very difficult, and the third question needs to be proved by some classic types! I try to scale with basic inequality, but if the scaling proves to be true, there is a premise that I, J and K must be arithmetic progression, so this method won't work ... Please continue to answer the last question of this question!