High school series, 200 bounty

This is authoritative information???

It is clear that the "authoritative sources" here are wrong!!

Analysis: We consist of binary linear equations.

y-x=mxy=n

Knowing that there is no solution to the equation, then δ< 0 to get m 2 + 4n < 0 At the same time, as long as m n satisfies this condition, the equation is unsolved.

For example, take m=0, n=-2 and get a(n+2)=m*a(n+1)+n*a(n).

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

Consider a(1)=0 and a(2)=2Rule.

a(1)=a(3)=a(5)=a(7)=...=0a(2)=2,a(4)=-4,a(6)=8,a(8)=-16 ..

This sequence equation has no solution, and it is clearly not a periodic sequence. So authoritative information is not authoritative!!

Addendum: Of course, if you have a positive number of m n here, there must be δ=m 2 + 4n >0 i.e. the equation has a solution;

Upstairs with the "difference equation" solution, it seems that he took m n as a positive integer to arrive at such an answer (I am not sure about it because I was not good at math in college).

Guys, if you're wrong, just warn me!

Reply to mvgt:

Yes, thank you very much. I've read your answer, and I know what you mean! ^_

This thing should be able to compete in high school students, isn't it a first-order constant coefficient recursive sequence, it's very simple, first get the eigenroots x1, x2, and then because x1 n, x2 n all satisfy the recursive formula, but not necessarily meet the initial term, so just combine them linearly!

I also took the number test today.

But I didn't have the heart to do it.

Excuse me. It's too much trouble, really.

My brain was dizzy in the morning.

You're a bit of a strong man, I'll obey you.

You will understand it when you study advanced mathematics.

Haha, I'm confused as a math department, this question is good.

Hehe, I really don't remember at all.

What I learned is now eaten with my meal.

If you haven't figured it out yet, I advise you to stop working hard, how tired you are.

Strongman, I'm 3 years older and can't understand it.

I'm sorry, I really don't.

Hehe, I really don't remember at all.

I really suspect that I studied ...... in vain for four years in college：

Don't rely on others to study on your own, the results calculated will be of great help to your academic performance in the future.

I spent three years in high school and three years in college, a total of six years.

Previously I would have done it, but, now I forgot.

After graduating for many years, I forgot all about it.

I thought I could, but ... Alas!!!

SN 2=3N 2AN+S(N-1) 2,AN≠0, SN 2-S 2=AN[SN+S]=3N 2*AN, SN+S=3N 2,S+SN=3(N+1) 2, subtract to obtain A+AN=3(2N+1),A-3(N+1)=-(AN-3N),AN-3N=(-1) (N-1)*(A-3), AN=3N+(-1) (N-1)*(A-3), B bn=E [A-AN].

e^[3(n+2)+(1)^(n+1)*(a-3)-3n-(-1)^(n-1)*(a-3)]

e 6, is a constant.

2)a-an=3-2(-1)^(n-1) *a-3)>0,|a-3|<3 2,3 2-bn] [a-an] = e n, where n is between an and a, in the condition of 2), an, n, kn.

The foot marker is a series of equal differences.

If the tolerance d=3, there are (99-3) 3+1=33.

in a series of equal differences.

A1, An and D are known

then n=(an-a1) d+1

Then I'll just write how to prove that "a sequence is a subsequence of a sequence".

If SK-1=(M-1)A1, then BK=AM can be proved (the reverse of certificate 1 is sufficient).

Let k be greater than 2sk=b1*(1-q k) (1-q)=a1*(1-q k) (1-q).

then bk=a (1-q k) (1-q).

Since A1 is a positive integer, SK greater than B1+B2+B3 shows that (1-q K) (1-q) is a positive integer greater than 2.

then the series is a subseries of the series.

Well, if I'm not clear enough, you can send me another message to ask me.

If I'm mistaken, please let me know and I'll think again. :-d

If SK-1=(M-1)A1, then BK=AM can be proved (the reverse of certificate 1 is sufficient).

Let k be greater than 2sk=b1*(1-q k) (1-q)=a1*(1-q k) (1-q).

then bk=a (1-q k) (1-q).

Since A1 is a positive integer, SK greater than B1+B2+B3 shows that (1-q K) (1-q) is a positive integer greater than 2.

then the series is a subseries of the series.

The derivative of a straight line is y'=1

The derivative of the curve is y'=1/x

Because of tangent, 1 x=1 x=1 is substituted for y=lnx, so this point is (1 0).

Substitute the linear equation a=-1

Select a method Using image analysis, the kernel of the AN image is vertically a straight line, and the BN image is a large long curve of the addition and modification of the liquid of the index function of the buried god, and because a1=b1 a4=b4 then there is b2 a2 b3 a3 according to the image, so s4 t4 selects a, method 2, and the special value method sets a1=1, d=1, then b1=1 b4=4 can also find s4 t4