The sum of the first n terms of the number series an is known to be Sn

Because 2sn=(n+2)an-1 n is an arbitrary positive integer.

2s(n-1)=(n+1)a(n-1)-12an=(n+2)an-(n+1)a(n-1)nan-(n+1)a(n-1)=0

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

2s1=3a1-1

s1=a1a1=1

an/a1=an/a(n-1)*a(n-1)/a(n-2)..a2/a1

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

an=(n+1)/2

2) tn=1/(a1*a2)+1/(a2*a3)+1/(a3*a4)…+1/(an*an+1)

4/2*3+4/3*4+..4/(n+1)(n+2)4(1/2-1/3+1/3-1/4...1/(n+1)-1/(n+2))

4(1/2-1/(n+2))

2n/(n+2)

Summary. Hello, glad to answer for you.

Test Points: Concepts and simple notation of number sequences.

Special Topics: Point Columns, Recursive Sequences, and Mathematical Induction.

Analysis: The general term formula of the series can be found according to the formula of the sum of the first n terms of the series given in the question, and the answer can be obtained by substituting n=6 into the general term formula

Answer: Solution: s6-s5=

So a6 = so the answer is:

Comments: This question examines the basic knowledge of the number series, tests the student's calculation ability, and should carefully review the question when solving the problem, answer it carefully, avoid mistakes, and belong to the mid-range question

Knowing that the sum of the first n terms of the sequence {an} is sn, and sn 2an 1, then a6 Hello, glad to answer for you. Test Points: Concepts and Simple Notation of Number Sequences:

Analysis of point column, recursive sequence and mathematical induction: Shan Xun can judge and calculate the general term formula of the number series according to the formula of the sum of the first n terms of the number series given in the question, and substitute n=6 into the general term formula to get the answer Answer: Solution:

s6-s5=2 66+1-2 55+1=121, so a6=121;Therefore, the answer is: 121 Comments: This question examines the basic knowledge of the number series, tests the student's calculation ability, and when solving the problem, it is necessary to carefully review the question and answer it carefully to avoid mistakes and digging, which belongs to the mid-range question

For the equation for the difference series: an=a1+(n-1)d;

The sum of the first n terms is sn=a1*n+(square of n-n)*d;

Because sn=2n-3;

So the two equations are corresponding, and the solution is:

a1=-1 d=2

So a4+a5+a6+a7=a1+3d+a1+4d+a1+5d+a1+6d=32

Solution 1 (General Solution):

a4+a5+a6+a7

a1+a2+a3+a4+a5+a6+a7)-(a1+a2+a3)

s7-s3＝(14-3)-(6-3)＝8；

Solution 2 (Special Solution):

Derived from known s[n] 2n-3.

a[n]＝s[n]-s[n-1]

2n-3)-[2(n-1)-3] 2, so a4+a5+a6+a7 2+2+2+2 8

A proportional series of 3.

an=(2/3)[∑n^2+3∑n+∑2]

an=(2/3)(n+2)(n+1)

bn=an/3)(n+1)

bn-b(n-1)=2 punch a(n-2)=(n+1) a1=4 2)n(n+1)+2n].

2n/，o(∩_o.;a1=(n+2) pronouncing emptiness (n-1).

na(n-1)/.1)∵a1=4;3*2)=(n+2)(n+1)/.n*(1/3)(n+2)(n+1)

a2/3)[(1/2

an/(n+1)=(2/9)(n^2+6n+11);3)(n^2+3n+2)

sn=(2/3)(n+2)

b(n-1)=(2/3

The common ratio is 2 6)n(n+1)(2n+1)+(3, I hope it will help you to be blind, (n+1)an+1=(n+3)an

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

q 2-3q+2=0 gives q=1 or q=2

2(a3+2)=a2+a4 gives 2(a1·q 2+2)=a1·q+a1·q 3

When q=1, there is no solution.

When q=2, the formula is 8A1+4=10A1, i.e., A1=2, then an=2·2 (n-1)=2 n

2) Solution: an·sn=2n 2(n+1).

Assuming that there is a series that satisfies the conditions, let the tolerance of this series be d

Method 1: [a1 (n 1)d] 2n2(n 1), i.e. n2 n

2n2 2n to any n n* constant true, then.

Solve or at this time an 2n or an 2n

Therefore there is a series of equal differences, such that for any nn*, there is an·sn 2n2(n 1), where an 2n or an 2n

(1) A series is a proportional series with a common ratio of 4.

sn+1=(s1+1)*4^n-1=(a1+1)*4^n-1=4^n(n>=1)

sn=4^n-1

When n>=2.

sn-1=4^n-1-1

an=sn-sn-1=3*4 n-1 brings in n=1 a1=3, which is a proportional series with 3 as the first term and 4 as the common ratio.

2) What is the problem?

Try to solve it yourself.

s(n+1)=2sn+1

s(n+1)+1=2(sn+1)

So there is: s2+1=2(s1+1).

s3+1=2(s2+1)

.s(n-1)+1=2(s(n-2)+1)sn+1=2(s(n-1)+1)

Multiply n-1 formulas to obtain: sn+1=2 (n-1)(s1+1), that is, sn=2 n-1(s1=a1)

From the formula of sn, it can be seen that an is a proportional series with a common ratio of 2. So an=2 (n-1).

s(n+1)=2sn+1 a2+a1=2a1+1 a2=a1+1 a2=1+1=2

sn=2s(n-1)+1 n>=2

Subtraction: an=2a(n-1).

an/a(n-1)=2

Because: a2 a1=2 1=2, for an a(n-1)=2 is true.

Therefore, an is a proportional series with 1 as the first and 2 as the common ratio.

an=2^(n-1)

s(n+1)=2sn+1

s(n+1)+1=2(sn+1)

Therefore, sn+1 is a proportional series with 2 as the first term, and the common ratio is 2, so sn+1=2 n

So an=s(n+1)-sn=2 n-2 (n-1)=2 (n-1).

Match the ratio to find sn, +1 on both sides, and find that it is an equal ratio series, which can be solved.

s(n+1)=2sn+1---1)sn=2s(n-1)+1---2)Eq. (1)-(2): a(n+1)=s(n+1)-sn=(2sn+1)-(2s(n-1)+1)=2(sn-s(n-1))=2an

Therefore, an is an equal proportional series of a1=1 and the common ratio q=2, and its general formula is: an=a1*q (n-1)=2 (n-1)

It is known that the sum of the first n terms of the sequence {an} is sn, and sn 2an 1, then a6