A high school math puzzle! Is it okay to find an without mathematical induction?

a(n+1)=(n-1)an/(n-an)a(n+1)/(n-1)=an/(n-an)n-1)/a(n+1)=n/an-1

1/(na(n+1))=1/((n-1)an)-1/n(n-1)1/(na(n+1))-1/((n-1)an)=-1/(n-1)n=-(1/(n-1)-1/n)

At n 2, there is:

1/((n-1)an)=1/((n-1)an)-1/((n-2)a(n-1))+1/1/((3-1)a3)-1/((2-1)a2)+1/((2-1)a2)

1/(n-2)-1/(n-1)+1/(n-3)-1/(n-2)+.1-1/2)+1/((2-1)a2)

1-1/(n-1)+4

3+1/(n-1)

an=1 (3n-2).

When n = 1, a1 = 1 is also satisfied.

Thus an=1 (3n-2)

Try subtracting or superimposing the graves by multiplying them, paying attention to the feasible domains.

This question is all said to ask you to find s1, s2, s3, s4, and then guess, what other method do you want to use?

It's not necessary, just calculate it by hand, s1-s4 and then guess (it's best to use the ordinary square skin method to make it) After all, I didn't let you prove Duan Yeming.

<> Hysan wandering hope is helpful to you, digging such as.

Don't you use mathematical induction?

Solution: (2) is composed of 4sn=an +2an

The result is: 4s(n-1)=a(n-1) +2a(n-1) where n 2 and n is an integer.

Subtract the two formulas to obtain:

4an=an -a(n-1) +2an-2a(n-1), i.e., an -a(n-1) -2an-2a(n-1)=0 factor the left side of the equation to :

an+a(n-1)][an-a(n-1)-2]=0 is positive by all, so an+a(n-1)>0

Therefore, there must be: an-a(n-1)-2=0

i.e. an=a(n-1)+2

Therefore, if we take 2 as the first term and 2 as the tolerance, then an=2+2(n-1)=2n

3)1/a1²+1/a2²+.1/an²=1/2²+1/4²+.1/(2n)²<1/2²+1/4·2+1/6·4+..

1/2n(2n-2)=1/4+[1/2-1/4+1/4-1/6+..1/(2n-4)-1/(2n-2)+1/(2n-2)-1/2n]/2

1/4+1/4-1/4n=1/2-1/4n∵n≥1

1/a1²+1/a2²+.1/an²<1/4+1/4-1/4n<1/2

1.a1=2 is not verbose.

2.Because a(n)=s(n)-s(n-1).

4s(n)=a(n)^2+2*a(n)

4s(n-1)=a(n-1)^2+2*a(n-1)

Subtract 4*a(n) = a(n) 2+2*a(n) -a(n-1) 2 - 2*a(n-1).

a(n-1)^2+2*a(n-1) +1 = a(n)^2-2*a(n) +1

a(n-1)+1)^2=(a(n) -1)^2

Probability 1 a(n-1)+1=a(n)-1

Possibility 2 a(n-1)+1=1-a(n) is a(n)+a(n-1)=0 Obviously, for a series of positive numbers, a1=2 is positive, and a2 is negative, so this possibility is ruled out.

So probability 1 is correct, a(n)-a(n-1)=2

n is taken from 2 to n, and there are a total of (n-1) formulas.

a2-a1=2

a3-a2=2

.a(n)-a(n-1)=2

All add up, and the front and back items are eliminated.

a(n)-a1=2(n-1)

a(n)=2n

Finished 3The scaling method proves that the inequality is actually a convergent series, and the series converges at 1 n.

Multiply the denominator 2 to the right to prove it.

1/n²<2, (n=1,2,..n)

Here's the point. 1/n² =1+1/(2*2)+ 1/(3*3)+.1/(n*n)

If the denominator is smaller, then the fraction will be larger.

1 n < 1 n(n-1), and 1 n(n-1)=1 (n-1)+1 n

So 1 n(n-1) = 1+1 (2*1) + 1 (3*2) +1 (n*(n-1)).

1+1-1/2+1/2-1/3+1/3-..1/(n-1)-1/n

2-1/n < 2

So, 1 n < 1 n(n-1) <2 are true for all positive integers n.