In a series of numbers an, where each term is positive, the first n terms and sn satisfy sn 1 2 an 1

If an = root number n - root number (n-1).

When n, a1 = 1 and a2 = root number 2-1 is clearly true.

Suppose that when n=k, s(k)=1 2 (ak-1 ak) also holds, and when n=k+1, s(k+1)=s(k)+a(k+1)=1 2 (root number k - root number (k-1) + root number k + root number (k-1)) + root number (k+1) - root number (k)).

The root number (k+1)=1 2(a(k+1)-1 a(k+1)) is established. Thus, for any n, an = root number n - root number (n-1) are true, and it is proven.

1) When n=1 sn=an, so sn=1 2(sn+1 sn), the solution is sn=1 or sn=-1 (rounded), and the conclusion is valid.

2) Suppose that when n = k, sk = root number k (k is an integer greater than 1).

then s(k+1)=1 2(a(k+1)+1 a(k+1))=1 2(s(k+1)-sk+1 (s(k+1)-sk)).

Substituting SK=root:K=K=1+root=1 (S(K+1)-root:K)

The solution is s(k+1) = root number (k+1) or s(k+1) = - root number (k+1) (rounded).

Therefore, when n=k+1 is the conclusion is also true.

To sum up, sn=root number n

n=1 is clearly true.

Assuming n=k, then when n=k+1, sk+a(k+1)=(a(k+1)+1 a(k+1)) 2

sk=(-a(k+1)+1 a(k+1)) 2a(k+1) 2+2*a(k+1)*sk-1=0 substitute sk=sqrt(k) to get a(k+1)=sqrt(k+1)-sqrt(k).

So s(k+1) = sk + a (k + 1) = sqrt (k + 1).

2sn=(an+1)*an=an^2+an2s(n+1)=a(n+1)^2+a(n+1)2s(n+1)-2sn=2a(n+1)=[a(n+1)^2+a(n+1)]-an^2+an)

The results are: a(n+1) 2-a(n+1)-an 2-an=0, that is, [a(n+1) 2-an 2]=[a(n+1)+an], [a(n+1)-an]*[a(n+1)+an]=a(n+1)+an

Since any an>0, a(n+1)+an on both sides of the equation can be eliminated, so there is a(n+1)-an=1

2sn=an 2+an, then 2a1=a1 2+a1,a1=1an=1+(n-1)*d=1+(n-1)*1=n, the process may be a bit troublesome.

6sn = (an+1) (an+2), so that n=1 then a1=1 (give up the reed) or a1=2

6sn-1=(a(n-1)+1)(a(n-1)+2) subtract [an+a(n-1)] [an-a(n-1)-3] = 0an-a(n-1)-3=0

an-a（n-1）=3

Equal grinding difference series.

an=2+3（n-1）=3n-1

sn=n（3n+1）/2

When n>=2, s[n]=1 4 * a[n]+1) 2; s[n-1]=1/4 * a[n-1]+1)^2

Subtract the two formulas to obtain a[n]=1 4 * a[n] 2+2a[n]-a[n-1] 2-2a[n-1])).

Simplification yields a[n] 2-a[n-1] 2=2a[n]+2a[n-1] and yields a[n]-a[n-1]=2, so it is a series of equal differences. The first term is 1, and the tolerance is 2a[n]=2n-1

The second step is not difficult, but it is more cumbersome to write it out.

The answer is tn=3-(2n+3) (2 n).

1) Because sn -(n +n-3)sn-3(n +n)=0, (sn+3)[sn-(n +n)]=0, because the terms of the sequence an are all positive numbers.

So sn-(n +n)=0

sn=n²+n

a1=s1=2

2) an=sn-s(n-1)=2n(n 2) When n=1, a1=2 1=2, so a1 conforms to the general formula.

Therefore, the general formula for the trouser clearance sequence {an} is an=2n

3）1/[a1(a1+2)]

1/[a2(a2+2)]+1/[an(an+2)]1/(4×6)+…+1 [an(an+2)]1 2 [1 2-1 4+1 4-1 6+....+1/an-1/(an+2)]

1/2×[1/2-1/(an+2)]

n/(4n+4)