The sequence a 0 1, a 1 1, a n a n 1 a n 2, find a n .

This sequence is the legendary Fibonacci sequence.

It has all the information you need.

In the sequence, a 1 = 1, a n+1 = a n +1, is the first term is 1, the tolerance is 1, and the difference is wide, a n = 1 + (n-1) 1 = n, a 100 = 100

So the answer is: 100

An is the sum of proportional sequences.

a=1an=1+1+……1=n

a≠1 then an=1+a+......a^(n-1)=1*(1-a^n)/(1-a)

1) Proof of Matching: by a

1 with. an+1an

anan +1

Anana: So, right? n∈n

an+1an is the number of burns in the permanent hall, therefore.

an is the series of equal difference Peiqin;

2) Obtained from (1). ana

2(n-1)=2n-1bn

anan+1

2n-1)(2n+1)

2n-12n+1 so. sn

bbbn2n-1

2n+12n+1n

2n+1 by. sn

Namely. n2n+1

Get. n So satisfied. sn

The smallest positive integer of n=503

With the following approximate general term formula:

(1) When n is an even number, let n=2k, then k=n 2sn=1 -2 +3 -4 +....2k-1)²-2k)²=(1-2)(1+2)+(3-4)(3+4)+…2k-1-2k)(2k-1+2k)

1-2-3-4-……2k-1)-2k=-(2k+1)*2k/2

k(2k+1)

n(n+1)/2

2) When n is odd, let n=2k-1, then k=(n+1) 2sn=1 -2 +3 -4 +....2k-3)²-2k-2)²+2k-1)²

1-2)(1+2)+(3-4)(3+4)+…2k-3-2k+2)(2k-3+2k-2)+(2k-1)²

1-2-3-4-……2k-3)-(2k-2)+(2k-1)²=-(2k-1)*(2k-2)/2+(2k-1)²=k(2k-1)

n(n+1)/2

In summary, sn=(-1) (n+1)*n(n+1)2

Solution: Let the common ratio of the proportional series be q, then there is the following relationship between it and sn, s2n, and s3n:

sn, s2n-sn, s3n-s2n are proportional sequences, and the common ratio is q n

Proof : Prove a more general formula first. In a proportional series, an=a1q (n-1).

am=a1q^(m-1)

Divide the two formulas to give an am=q (n-m), an=amq (n-m)

s2n=a1+a2+..an+a(n+1)+a(n+2)+.a2n

sn+(a1q^n+a2q^n+..anq^n)=sn+(a1+a2+..an)q^n=sn+snq^n

s2n-sn)/sn=q^n.

In the same way, s3n=s2n+[a(2n+1)+a(2n+2)+.a3n]

s2n+[a(n+1)q^n+a(n+2)q^n+..a2nq^n)

s2n+[a(n+1)+a(n+2)+.a2n]q^n

s2n+[s2n-sn}q^n.

s3n-s2n)/(s2n-sn)=q^n.

s2n-sn)/sn=(s3n-s2n)/(s2n-sn).i.e. (s2n-sn) 2=sn(s3n-s2n)Evidence.

Solution: In a sequence, a[n+2]-a[n]=1+(-1) n When n is odd: a[n+2]-a[n]=0 When n is even: a[n+2]-a[n]=2

a[1]=2，a[2]=2

In a sequence: its odd terms form a constant subseries with a constant of 2.

Its even terms form a series of equal subnumbers with a starting term of 2 and a tolerance of 2: a[100]=a[2]+2*=100

s[100]=(a[1]+a[3]+.a[99])+a[2]+a[4]+.a[100])

(99-1)/2+1]a[1]+[100-2)/2+1](a[2]+a[100])/2

Hello, a1=1

a2=2a3-a1=1+(-1)^1=0 --a3=a1=1a4-a2=1+1=2 --a4=2+a2=4a5-a3=0 --a5=a3=1

a6-a4=2 --a6=2 a4=6 shows that the odd term of this series is always equal to 1, and the even term = n

So s100=a1+a3+a5+.a99+a2+a4+a6+..a100

Evidence: (1) Because.

an+1anan

anan so. an+1

Ann n.

An is a series of equal differences

From this we can get it, 1an

a(n-1) (1)=-n so. a

nnn-1nn∈n

2) by conditions. bn

ann can know that when n=2k,b

n0；When n=2k-1, b

n0, k n order. b

Nann stared at Bi. bn+1

bnnn+1

n+1n-1nnn

nn+1n-1n

nn5n(n+1) dang-n

5＞0?n infiltrate 2, |b

n+1＞|bn

The same can be obtained, when -n

5＜0?n 3, |b

n+1＜|bn

That is, the number series is increasing when n=1,2,3; n 4, decreasing; Namely. b is the largest term in the series

However, because {b

The odd terms of n are -|bn, therefore. b

for the series {bn

the smallest term; Whereas. b

b so bb so b is a sequence {bn

The maximum term is for any positive integer m, n, bnbmbb

Sequence. bnannn∈n

is a "domain convergence sequence".

1）a6，a12，a

20；…(6 points).

2) The rotten conjecture. an

n(n+1)， n∈

n Prove by mathematical induction:

1) When n=1, a

1 (1+1)=2, the proposition is true.

2) Suppose that the proposition is true when n=k(k1), i.e., akk(k+1).

Then when n=k+1, a

k+12(k+1

ak2(k+1

k(k+1)

k+1)(2k+2-k)

k+1)(k+2)

k+1)[(k+1)+1]

So, when n=k+1 the proposition is also true.

From 1), 2) we can get Sun Chan's integer n,ann(n+1), n for any regular dust

n…(12 points).