How to find the value of a sequence an?

Use the accumulation method. Because 1 [(2n-1)(2n+1)]=[1 (2n-1)-1 (2n+1)] 2, so.

a1=3a2=a1+(1-1/3)/2

a3=a2+(1/3-1/5)/2

a4=a3+(1/5-1/7)/2

a5=a4+(1/7-1/9)/2

an=a(n-1)+[1 (2n-3)-1 (2n-1)] 2 The above equations are added separately on both sides, and the dislocation is used to cancel each other.

an=3+1/2-1/[2(2n-1)]=3+(n-1)/(2n-1)。

The value of the first term n = first term + (number of terms - 1) tolerance.

an=am+(n-m)d, if a certain term am is known, you can list the formulas related to d to solve an.

For example, a10=a4+6d or a3=a7-4d.

The sum of the first n terms sn=first term n + number of terms (number of terms - 1) tolerance 2.

Tolerance d=(an-a1) (n-1) (where n is greater than or equal to 2 and n is a positive integer).

Number of Items = (Last Item - First Term) Tolerance + 1.

Last term = first term + (number of terms - 1) tolerance.

When the number column is odd, the sum of the first n terms = the number of intermediate blind terms.

The number column is an even number of terms, and the sum of the first n terms = (the sum of the first and last terms) 2.

Note.

The difference series is a common series, which can be expressed by AP, if a series of numbers from the second term, the difference between each term and its previous term is equal to the same constant, this series is called the difference series, and this constant is called the tolerance of the difference series, and the tolerance is often represented by the letter D.

For example: 1, 3, 5, 7, 9 ......（2n-1)。The general formula for the equal difference series is:

an=a1+(n-1)d。The first n terms and formulas are: sn=n*a1+n(n-1)d2 or sn=n(a1+an)2[2].

Note: The above integers.

a(n+1)-an=2n+3

So. an-a(n-1)=2(n-1)+3a(n-1)-a(n-2)=2(n-2)+3 Yanhui is several years old....a2-a1=2*1+3

Add. an-a1=2[1+2+……Rough answer + (n-1)] + 3 (n-1) = n 2 + 2n-3

a1=3, so an=n2+2n

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a(n+1)/(n+1) -an/n=½ⁿ=½ⁿ⁻a(n+1)/(n+1)+½=an/n +½a1/1 +1=1/1 +1=2

A sequence is a constant sequence of numbers where all terms are 2.

an/n +½=2

an=n·(2ⁿ-1)/2ⁿ⁻¹

When n=1, a1=1· (2-1) 1=1, a1=1 also satisfies the general formula of the expression sequence, which is an=n· (2 -1) 2

a2=(3/1)*a1

a3=(4/2)*a2

a4=(5/3)*a3

a5=(6/4)*a4

an=((n+1) (n-1))*a(n-1) The above n-1 equations are multiplied and simplified.

an=((3/1)*(4/2)*5/3* *n+1)/(n-1))*a1

n*(n+1)/2

Let an=tan then an+1=tan( +4), so it is a periodic series.

From the title: a1=1004 1005,a2=-1 2009,a3=-1005 1004,a4=2009,a5=1004 1005

So ak=AK+4 (k is a positive integer), a2008=a4=2009

a(n+1)=(an-1) (an+1),a3=(a2-1) (a2+1)=-1005 1004,a2=-1 2009

a2=(a1-1) (a1+1)=-1 2009, a1=-1004 1005

a4=(a3-1)/(a3+1)=-2009a5=(a4-1)/(a4+1)=1005/1004a6=(a5-1)/(a5+1)=1/2009a7=(a6-1)/(a6+1)=-1004/1005...You can get six in a cycle.

Therefore a2008=a4=-2009

From the meaning of the title, it can be seen that (an+1an+2) (an+1an)=q q 0(anan+1)+(an+1an+2) (an+2an+3), that is, (anan+1)+q(anan+1) q 2(anan+1) because it is a positive number series, so both sides are divided by (anan+1) inequality at the same time, and the direction is unchanged to obtain q 2-q-1 0 combined with q 0 to get 0 q (1+root number 5) 2

ps: The picture can't be uploaded, so I'll just look at it.