(1) I don't know what to prove.

(2) Prove that √(2n- 1)≤an≤√(3n-2)

(3) Find a positive integer m so that |a20 17-m| is the minimum.

(2)

It is proved that the inequality holds when n= 1, 2, 3, 4. Assuming that the inequality holds when n=N, that is √(2N- 1)≤aN≤√(3N-2), then 2n- 1 ≤ an 2 ≤ 3n-2.

Then when n=N+ 1, 2 (n+ 1)- 1

So √ [2 (n+1)-1] ≤ A (n+1) ≤√ [3 (n+1)-2]

So when n=N+ 1, the inequality also holds. That is, for any positive integer n, there is √(2n- 1)≤an≤√(3n-2).

(3)

According to (2), √ 3969 = 63

For convenience, we call a20 17 as going back to traverse a20 16, a20 15, ..., an as downlink, while going forward to traverse a20 18, a20 19, ..., ak as uplink.

1/78<a 20 17-a 20 16 = 1/a 20 16 & lt； 1/63， 1/78 & lt； a 20 18-a 20 17 = 1/a 20 17 & lt； 1/63

Then the above two formulas indicate that the falling time should not exceed 78 times at most, and the value of an will be reduced by1compared with a20 17; On the uplink, the value of ak needs to be at least 63 times higher than that of a20 17 1, because the decline speed of an will be faster and faster on the downlink, and the increase speed will be slower and slower on the uplink.

Now let's look at the situation that a(20 17-78)=a 1939 and a(20 17+63)=a2080.

62 & lt√3877≤a 1939≤√58 15 & lt； 77，64 & lt； √4 159≤a 2080≤√6238 & lt； 79

4033≤a20 17^2≤6049

4033=3n-2，n = 1345； 6049=2n- 1，n=3025，3025- 1345= 1680

Then 2689 ≤ A 1345 2 ≤ 4033, 6049 ≤ A3025 2 ≤ 9073, 6049-2689 = 3360 = 1680 * 2, and the lower limit is not counted.

269 1≤a 1346^2≤4036,6047≤a3024^2≤9070

1/4033+2≤a 1346^2-a 1345^2= 1/a 1345^2+2≤ 1/2689+2

1/9070+2≤a3025^2-a3024^2= 1/a3024^2+2≤ 1/6047+2

20 17- 1345=672, and the upper limit is 4033+672 * 2 = 5377,672/4033.

3025-20 17= 1008, and the lower limit is 6049- 1008*2=4033.

3025- 1345= 1680,4033+ 1680*2=7393,7393- 1008*2=5377

2689=3n-2,n=897, 1793≤a897^2≤2689, 1795≤a898^2≤2692，

2+ 1/2689≤a898^2-a897^2= 1/a897^2+2≤2+ 1/ 1793

2017-897 = 1 120,2689+1120 * 2 = 4929 = a 20172 upper limit,/kloc-0. mistake

1793=3n-2,n=599, 1 197≤a599^2≤ 1795，

2+ 1/ 1795≤a600^2-a599^2=2+ 1/a599^2≤2+ 1/ 1 197

2017-599 =1418,1795+1418 * 2 = 4633 = a2017 2 upper limit. mistake

1 197+ 1=3n-2,n=400,799≤a400^2≤ 1 198，

2+ 1/ 1 198≤a40 1^2-a400^2=2+ 1/a400^2≤2+ 1/799

2017-400 =1617,1201+1617 * 2 = 4435 = a20/kloc. mistake

799=3n-2,n=267,533≤a267^2≤799，

2+ 1/799≤a268^2-a267^2=2+ 1/a267^2≤2+ 1/533

2017-267 =1750,799+1750 * 2 = 4299 = a 20172 upper limit, 1750/799.

533+ 1=3n-2,n= 179,357≤a 179^2≤535，

2+ 1/535≤a268^2-a267^2=2+ 1/a267^2≤2+ 1/357

2017-179 =1750,535+1838 * 2 = 421= a 20172 upper limit, mistake

359- 1=3n-2,n= 120,239≤a 120^2≤358，

2+ 1/358≤a 12 1^2-a 120^2=2+ 1/a 120^2≤2+ 1/239

2017-120 =1750,358+1897 * 2 = 4152 = a 20172 upper limit, 4

We can finally finish here, because the maximum error of the upper limit of A2017 24152 is less than 64.5 even if the square root of 8 is added.

The lower limit of A2017 2 4033 is greater than 63.5, so m=64.