The known sequence an full a1 1, an a1 1 2a2 1 3a3 1 n 1an 1 n 2, n belongs to n an 2006, n

a2=a1=1

n>=3.

an+1=a1+1/2a2+..The subtraction of 1 n-1an-1+1 nan gives an+1-an=1 nan

i.e. an+1=n+1 nan

i.e. an+1 an=n+1 n

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

n+1 2 is an+1=n+1 2

i.e. an=n 2

an=2006, I'll do it. Thanks, ha

n=4012

I will, you wait.

Add all the left sides of the above equation and all the right sides to get it.

an+1-a2=an/n-a1

With known conditions, a2=a1=1 can be obtained

So an+1=an n

So an+1 an=1 n

So there is an an-1=1 (n-1).

Multiply the left by the left and the right by the zip code to get an=1 (n-1)(n-2)(n-3)(n-4) 1

Bring in an=2006

You can come out on the left.

But your 2006 should be 1 2006.

n can be found

a2=a1/1=1

a2=a1=1

n>=3.

an+1=a1+1/2a2+..1/n-1an-1+1/nan

Subtract the two formulas to give an+1-an=1 nan

i.e. an+1=n+1 nan

i.e. an+1 an=n+1 n

an+1/a2=(an+1/an)(an/an-1)..a3/a2)

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

i.e. an+1=n+1 2

i.e. an=n 2

an=2006,n=4012

n=2006

Because an=2006

must be an integer.

Since a1=1, a2=1 (2-1)a(2-1)=1=a1, so an=1+(n-2) 2

And because an=2006

So n=4012

Greater than or equal to 2, n belongs to n).

1/a(n)=（-1）^n-2/a(n-1)

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

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

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

So the number column is proportional to the series, and the common ratio is q=-2

1/a(n+1)+1/a(n)=[1/a(2)+1/a(1)]q^(n-1)

a(1)=1/4,a(2)=-1/7

1/a(n+1)+1/a(n)=3*(-2)^(n-1)/28

1)^(n+1)-2/a(n)+1/a(n)=3*(-2)^(n-1)/28

a(n)=28/[(-1)^(n-1)(28-3*2^(n-1)]

1/b(n)=(1-3*2^(n-1)/28)^2

1/b(n+1)=(1-3*2^n/28)^2

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

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

So the number column is proportional to the series.

b(1)=16,b(2)=49

b(n)=16*(49/16)^(n-1)

s(n)=16(1-(16/49)^(n-1)/(1-16/49)=784(1-(16/49)^n)/33

First, find the general term formula of the series, which is a regular recursive relationship, and the interval terms are proportional.

Then divide the 2021 terms into odd and even terms, and sum them separately.

For reference, please smile.

Solution: a2=(1+a1) (1-a1)=(1+2) (1-2)=-3

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

a4=(1+a3) (1-a3)=(1-1 ruler base2) (1+1 2)=1 3

a5=(1+a4)/(1-a4)=(1-1/3)/(1+1/3)=1/2

a6=(1+a5)/(1-a5)=(1+1/2)/(1-1/2)=3

a7=(1+a6)/(1-a6)=(1+3)/(1-3)=-2

a8=(1+a7)/(1-a7)=(1-2)/(1+2)=-1/3

a9=(1+a8)/(1-a8)=(1-1/3)/(1+1/3)=1/2

Irregular spine: The first 4 terms of the number series are 2-3-1 21 3

Item 5 starts with 1 23-2-1 3 cycles every 4 cycles.

2013 4 = 503 remainder 1

a2013=1/2

a1×a2×a3×..a2013

a1×a2×a3×a4)×(a5×a6×a7×a8)×.a2009×a2010×a2011×a2012)×a2013

1×1×..1×a2013a2013

a2=2^2-1*2+1=3

a3=3^2-2*3+1=4

a4=4^2-3*4+1=5

a5=5^2-4*5+1=6

Guess an=n+1

The following is proved by mathematical induction.

From a1=2=1+1 to n=1, an=n+1 is true.

Let n=k(k is a positive integer) when an=n+1 becomes immediately ak=k+1, then when n=k+1, because a(n+1)=an -n*an+1, then a(k+1)=ak -k*(k+1)+1

k+1)²-k*(k+1)+1

k²+2k+1-k²-k+1

k+2 is summed up, an=n+1 is true.

From the known sequence, a[1]=1,a[n+1]=2a[n]-n +3n Note: [Inside is the subscript.

1) a[2]=4,a[3]=10

2) Let c[n]=a[n]+ n2+ n

It is known that a[1]=1

When n>=2, a[n]=2a[n-1]-n 2+5n-4

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

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

=-1, =1

3) c[n]=a[n]-n 2+n from (2).

c[1]=1, and n>=2 c[n]=2c[n-1].

So c[n]=2 (n-1).

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

b[n]=1/(a[n]+n-2^(n-1))=1/n^2

It can be proved that s[1]=1<5 3,s[2]=5 4<5 3,s[3]=49 36<60 36=5 3

s[4]=205/144<240/144=5/3 s[5]=5129/3600<6000/3600=5/3

When n>=6.

s[n]=s[5]+a[6]+.a[n]<5129/3600+1/(5*6)+1/(6*7)+.1 ((n-1)*n) Zoom in from item 6.

5129/3600+(1/5-1/6)+(1/6-1/7)+.1/(n-1)-1/n)

For all positive integers n,s[n]<5 3

This question should be enlarged from item 6.

Hope it helps you a little!

I have to say that the person who asked the question of this question is absolutely under-flat, and a simple split-term method actually has to be reduced to the fifth item to meet the meaning of the question, and simply increase the amount of calculation.

The third question is to use factorization to add up the re-splitting terms.

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

2an+2 n+1=an+1 divided by 2 n+1 on both sides.

an 2 n) +1=(an+1) (2 n+1) So the sequence is a series of equal differences with the first term being 1 and the tolerance being 1.

2. Let an 2 n=bn

Then bn=n so an=n x 2 n

sn= 1x2+2x2^2+3x2^3+..nx2^n

2sn=1x2^2+2x2^3+3x2^4+..nx2^n+1

2sn-sn= --2+2^2+2^3+….2^n)+nx2^n+1=sn

So sn= nx2 (n+1).

- 2^(n+1) +2 =(n-1)x2^(n+1) +2

The first one is divided by 2 to the nth power, and the second one finds the general term of an, and then calculates, the sn expression should be a proportional series plus an equal difference series. The first term is 1, and the tolerance is 1 -sn=[2(1-2 n)] (1-2)-n*2 (n+1).

Solution: (1) Because an=(1 2)a(n 1)-2 n, so a(n 1)=2an 2 (n 1), so a(n 1) 2 (n 1)-an 2n=[2an 2 (n 1)] 2 (n 1)-an 2 n=an 2 n 1-an 2 n=1, so the series is a series of equal differences with the first term 2 and the tolerance of 1.

a1=1

a2=a1=1

a3=a1+a2/2=1+1/2=3/2

Conjecture: When n>1, there is an=n 2

Proof: (1) When n=2, there is a2=2 2=1 true.

2) Let n=k, there is ak=k 2

Then when n=k+1, there is ak+1=a1+a2 2+a3 3+.ak-1/(k-1)+ak/k

And a1+a2 2+a3 3+...ak-1/(k-1)=ak∴ak+1=ak+ak/k=ak*(1+1/k)=k/2*(k+1)/k=(k+1)/2

That is, when n=k+1, there is ak+1=(k+1) 2 is true, which can be seen from (1)(2), and for n>1, an=n 2 is true.

an=1(n=1),an=n/2(n>1)