In the known sequence an, a1 2 and an an 1 n 1 n 1, then an

Solution: Considering that each term of an is a proportional recursive relationship with the previous term, it can be multiplied to simplify the general term:

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

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

4/[n(n+1)]

Hope it helps you to understand pull.

Is an-1 the n-1 term of an or (an)-1? If the second is listed in the division cross multiplication an=-n, if the first one is multiplied by quotient an=(a2 a1)*(a3 a2)*(a4 a3)*....an-1 an-2)*(an an-1)=2 [n(n+1)] should look like this.

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

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

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

an=(n-1/n+1)*(n-2/n)*(n-3/n-1)*…1/3*2

n = odd number.

an=(1/n+1)*(1/n)*4

n=even.

an=(n-1 n+1)*(1 n)*(1 n-1)*4 are all (1 n+1)*(1 n)*4

an=an-1/2an-1+1

Take the above equation to the reciprocal to get it.

1 an=(2an-1+1) an-1=1 an-1+2, so 1 an

1/an-1=2

So the number column is an equal difference series.

1/a1=1

Then 1 an = 1 + 2 (n-1) = 2n - 1

an=1/(2n-1)

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

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) When a=-7, an=1+1 [a+2(n-1)]=1+1 [-7+2(n-1)]=1+1 is a subtractive function.

When n=1, an takes the maximum value: 6 7. When n approaches infinity, the limit value of an is 1. That is, the maximum value is 6 7 and the minimum value is 1

2) From an=1+1 [a+2(n-1)], an a6, we get: 1+1 [a+2(n-1)] 1+1 [a+2(6-1)], i.e., n>=6.

Why does your step an=1+1 2 n-2-a 2 satisfy 5<2-a 2<6?How did you get here, it looks dizzy.

an-1 -an) an=(an- an-1) an-1, i.e. a(n-1) an-1=an a(n-1)-1, so an -a(n-1) = 0

an+a(n-1)][an-a(n-1)]=01. an+a(n-1)=0

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

an=a1=2 when n is odd and an=a2=12 when n is even an-a(n-1)=0

an=a(n-1)=.a2=a1

It is known that a1=2 a2=1

Therefore, the above formula does not hold.

To sum up: an=2 (n is an even number) or 1 (n is an odd number) I hope it can help you, I wish you progress in learning o( o

When n is an even number, an=2

When n is an odd number, an=1

a(n+1)=an +1/(n²+n)=an +1/[n(n+1)]=an +1/n -1/(n+1)

a(n+1) +1 (n+1)=an +1 na1+1 1=1 2 +1=3 2, and the sequence is a constant sequence of numbers where each item is 3 2.

an +1/n=3/2

an=3/2 -1/n

Let b[n]=a[n]-2n

Substituting the original formula gives b[n+1]=2b[n] (which can be checked) and then finds b[1].

If you get the b[n] general term, then you also get the a[n] general term.

a[n]=2^n+2n

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

sn>=an+2n^2

i.e. 2 (n+1)-2+n 2+n>=2 n+2n+2n 22 n>=n 2+n+2

Since 2 n and n 2+n+2 are both increasing functions on n>=1. And there are 2 5 > = 5 2 + 5 + 2

Therefore the smallest positive integer of n is n=5

(1)a(n+1)=2(an-n+1）

a(n+1)-2(n+1) = 2(an-2n)[a(n+1)-2(n+1)] (an-2n) =2=> is a proportional series.

2)(an-2n ) /(a1-2 )=2^(n-1)an = 2n+ 2^n

sn =a1+a2+..an

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

sn > an +2n^2

n(n+1) +2(2^n-1) >2n+ 2^n + 2n^2n^2+n -2 +2^(n+1) >2n^2+2n)+ 2^n2^n > n^2+n+2

The minimum value of a positive integer n = 6

A1=2>0, assuming that when n=k(k n*) and AK>0, then a(k+1)=AK+2AK>0

k is an arbitrary positive integer, so for any positive integer n, an>0a(n+1)=an +2an

1+A(N+1)=1+AN +2AN=(1+AN) LG[1+A(N+1)]=LG(1+AN) =2LG(1+AN)LG[1+A(N+1)] LG(1+AN)=2, which is the fixed value LG(1+A1)=LG(1+2)=LG3

The number series is an equal proportional series with LG3 as the first term and 2 as the common ratio.

lg(1+an)=lg3·2ⁿ⁻¹=lg3^(2ⁿ⁻¹1+an=3^(2ⁿ⁻¹

tn=(1+a1)(1+a2)..1+an)=3^(2⁰)·3^(2¹)·3^(2ⁿ⁻¹=3^(1+2+..2ⁿ⁻¹