A high one year number series, a high two math number series

a1(1-q^n)/(1-q)=1

1 a1*(1-1 q n) (1-q)=4 is divided by the two formulas.

a1*a1q (n-1)=1 4 i.e.

a1*an=1/4

Set a1*a2*a3*.an=p

then an*a(n-1)*a(n-2).a1=p is multiplied by the terms, [a1*an][a2*a(n-1)][a3a(n-2)].a(n-1)*a2][an*a1]=p^2

According to the nature of the proportional series, [a1*an]=[a2*a(n-1)]=[a3a(n-2)]=....=[a(n-1)*a2]=[an*a1]=1/4

So the above equation further becomes (1 4) n=p 2

Simplified to p=(1 2) n

That is, the original formula = (1 2) n

Generally speaking, the basic idea of finding the general term of a series of numbers is to convert it into the simplest series of equal differences and proportions to solve it.

As for the immobility method, it is a shortcut to find transformation.

So instead of this shortcut, the purpose of other methods is the same.

For example, if you don't have to move this question, then find a way to change it.

Generally, in the process of changing, the term with a(n) and the term with a(n+1) should be separated and the structure is the same, and the method of taking the reciprocal is often used.

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

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

Transform the same structure Try to adjust the left and right molecules to similar structures (adjusting the denominator is more difficult).

In this case, the pending coefficient is also used.

x+a(n+1)=x+1/(2-a(n))=2x+1+xa(n)) 2-a(n))

Then x+a(n+1) should be proportional to (2x+1+xa(n)) to be considered similar.

Then x (2x+1)=1 x gives x=-1

Then it is a(n+1)=1 (2-a(n)) minus 1 on both sides

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

The same structure appears, and for this wild song, the reciprocal is generally taken for processing, and the parameters are separated for the right.

Then it is 1 (a(n+1)-1)=(2-a(n)) a(n)-1)=1 (a(n)-1)-1

At this time, the equal difference series was born, and it was a equal difference series with a tolerance of -1, and you didn't give it to me. . . .

Then solve 1 (a(n)-1) and solve a(n).

In fact, looking back, the core of the immovable method is the same, except that the process of the immovable method is simple, and there is no need to think about it.

Solution: The entire equation is divided by a(n)*a(n+1).

Get 2 a(n) chong you order = 1 + 1 a (disturb n n + 1) 1 a (n + 1) -1 = 2 (1 a (n) - 1) let b (n) = 1 a (n + 1) -1, you can get.

b(n)} is a proportional series with 1 a(1)-1 as the first term and 2 as the common ratio.

1/a（n）-1=[1/a（1）-1]*2^(n-1)a(n)=1/

The condition of the problem is not enough, there must be a condition that can find a certain term, such as the value of a(1).

The general term is an=1+2+.n=n(n+1)/2=(n^2+n)/2

So the sum of the first n terms is sn=a1+a2+.an=(1^2+1)/2+(2^2+2)/2+..n^2+n)/2

(1^2+2^2+..n^2)+(1+2+..n)]/2

n(n+1)(2n+1)/6+n(n+1)/2]/2

n(n+1)(n+2)/6

The formula used above is 1 2 + 2 2 + 3 2 +...n 2=n(n+1)(2n+1) 6 This is how it comes.

Make use of the cubic variance formula.

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

n^3-(n-1)^3=2*n^2+(n-1)^2-n

The equations are added in full.

n^3-1^3=2*(2^2+3^2+..n^2)+[1^2+2^2+..n-1)^2]-(2+3+4+..n)

2*(1^2+2^2+3^2+..n^2)-2+[1^2+2^2+..n-1)^2+n^2]-n^2-(2+3+4+..n)

3*(1^2+2^2+3^2+..n^2)-2-n^2-(1+2+3+..n)+1

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

3(1^2+2^2+..n^2)=n^3+n^2+n(n+1)/2=(n/2)(2n^2+2n+n+1)

n/2)(n+1)(2n+1)

1^2+2^2+3^2+..n^2=n(n+1)(2n+1)/6

Solution: (1) If k=0 then the numerator an+2-an+1=0 column is a constant sequence, then an+1-an is also 0, the denominator = 0, contradictory, so k cannot be 0, that is, correct;

2) The equal difference series with a tolerance of 0 is not an equal difference ratio series, because the denominator is 0 at this time, which is contradictory. So wrong;

3) The proportional series with a common ratio of 1 is not a proportional series, and the denominator is 0 at this time, which is contradictory. So wrong;

4) The question says that there can be, so as long as you find a satisfying one, it means that it is right. And the number series 0 1 0 1 0 1 ......Apparently a series of proportional differences, so correct.

The first pair, because if k = 0, then the numerator is 0, that is, an+2=an+1, and an is a constant sequence, then the denominator is also 0, so k is not equal to 0. The fourth pair, e.g. 0,1,0,1... So a series of numbers will do.

Both the second and third can be illustrated by a constant column an=1 to illustrate the error.

does not exist. Solution: Let {an} be a series of equal differences, and the number {1 an} is also a series of equal differences.

2an=a(n-1)+a(n+1）① 2/an=1/a(n-1)+1/a(n+1）②

Simplified to obtain 2a(n-1)·a(n+1)=an[a(n-1)+a(n+1)].

That is, 4a(n-1)·a(n+1)=2an[a(n-1)+a(n+1)]=a(n-1)+a(n+1)].

Move the item. a(n-1)-a(n+1)]²0

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

then an is a series of equal differences with a tolerance equal to 0.

Solution: Using the fixed point method, since a(n+1)=(13an-25) (an+3) makes x=(13x-25) (x+3).

The solution yields x1=x2=5

Because x1 = x2 = 5

Then there is 1 (a(n+1)-x1)=1 (an-x1)+p in the equation x=(ax+b) (cx+d):

p=2c/(a+d) =1/8

So for the difference series, the tolerance is p=2c (a+d) =1 8(1) if a1=5, a2=(13 5-25) (5+3)=5, and the same way a3=5,......Guess ak=5, when n=k+1, a(k+1)=(13ak-25) (ak+3)=(13 5-25) (5+3)=5

Therefore, for n=k+1, it is also true.

So an=5

2) If a1=3, 1 (a1-5)=-1 2, so 1 (an-5)=-1 2+1 8(n-1)=(n-5) 8, i.e. an-5=8 (n-5).

The solution yields an=(5n-17) (n-5).

3) When a1=6, 1 (a1-5)=1

So 1 (an-5) = 1+1 8 (n-1) = (n+7) 8 i.e. an-5 = 8 (n+7).

The solution yields an=(5n+43) (n+7).

Why use the characteristic root method or the immovable point method to solve it, and use a knife to kill chickens. Tell you to present the denominator to an+1*(an+3)=(13an-25), and then use the equation idea.

Find the fixed point first, so that x=(13x-25) (x+3) solves x1=x2=5 and then n belongs to n

Therefore, a(n+1)-5=(13an-25) (an+3)-5=8(an-5) (an+3)...

1) If a1=5, it is easy to find a2=5

Guess ak=5, when n=k+1, a(k+1)=(13ak-25) (ak+3)=(13 5-25) (5+3)=5

Therefore, for n=k+1, it is also true.

So an=5

2) Let bn=1 (an-5), b1=1 (a1-5)=-1 2(*) be reduced to: b(n+1)-bn=1 8

For the difference series, it is easy to find bn=b1+1 8(n-1)=1 8n-5 8, so 1(an-5)=1 8n-5 8

Find an=(5n-17) (n-5).

3) When a1=6, the same is true for bn=b1+1 8(n-1), where b1=1 (a1-5)=1

So bn=1 8n+7 8

So 1 (an-5) = 1 8n + 7 8

So an=(5n+43) (n+7)(4)

（1）sn=2an-2，s（n+1）=2a（n+1）-2.

Subtract the two formulas to give a(n+1)=2a(n+1)-2an.

a(n+1)=2an, which is solved from the original formula and a1=2.

Therefore, the number series is an equal proportional series with the first term 2 and the common ratio of 2.

Thus an=a1q=2.

a（n+2）-a（n+1））/（a（n+1）-an）=2.

In summary, the general formula of the series is an=2, and it is a series of proportional differences.

2) If it is a series of equal differences, you may wish to set an=a1+(n-1)d.

Therefore, (a(n+2)-a(n+1)) (a(n+1)-an)=d d=1

To sum up, it must be a series of proportional differences.

3) Let an=2 +1, which is neither equal nor proportional.

a（n+2）-a（n+1））/（a（n+1）-an）=2.

Therefore, an is also a difference proportional series.

Assuming the common ratio is q, then.

a2=a1*q,a3=a1*q 2,a4=a3+(a3-a2)=a1(2q 2-q)so a1+a1(2q 2-q)=16,a1*q+a1*q 2=12 to get a1=1,q=3, or a1=16,q=1 2 so a1=1,a2=3,a3=9,a4=15 or a1=16,a2=8,a3=4, a4=0