1。1。2。3。5。8。13。21。34。55 What is the 30th number recursively?

Number 11 89

Number 12 144

Number 13 233

Number 14 377

Number 15 610

Number 16 987

Number 17 1597

Number 18 2584

Number 19 4181

Number 20 6765

Number 21 10946

Number 22 17711

Number 23 28657

Number 24 46368

Number 25 75025

Number 26 121393

Number 27 196418

Number 28 317811

Number 29 514229

Number 30 832040

It's a series of names with a lot of names.

Fibonacci series (Fibonacci sequence).

The operation is complicated, and you can see the Fibonacci sequence in the encyclopedia, and it's more detailed there.

The recursive formula for the Perbonacci sequence: f(n+2) = f(n+1) +f(n).

Numerator: Denominator.

Odd terms are negative, and even numbers are positive.

So (-1) n

The nth volume of stuffy stool posture leakage jujube rot is:

1)^n×[n×(n+1)]/3^n

1/3=-1/(1*3),-2/15=-2/(3*5), 3/35= 3/(5*7),-4/63=-4/(7*9), 5/99= 5/(9*11),.It is not difficult to see that the general argument formula is: an=[(1) (n-1)]*n [(2n-1)*(2n+1)] n>=2) when n=1 an=-1 3, so the next value should be n=6....

This sequence consists of two separate sequences.

The general formula for an even term is: a(n) =1/(n/2+1)

The general formula for odd-numbered terms is: a(n) =1-2n

Number 2023 is an odd-numbered term for:

a(2023) =1 - 2*2023 = 4045

Analysis: Odd digits: 0, -1 2, -2 3, -3 4....n-1)/n

Even digits: 1 3, 3 5, 5 7, 7 5,..2n-1)/(2n+1)

2017 is an odd number, 2017 2=

The 2017th number is: -1008 1009

84 1 8 3 10 is the (84th) number, and the 50th number is 1 8

Odd terms are negative, and even items are positive.

The molecule is ......

The denominator is ......

The nth term is: (-1) n n (n+1) n is a positive integer.

You can divide odd and even numbers to find two:

Odd number: nth number = (n+1) 2

Even number: nth number = -n (n+2).

Then the tenth number is -5 6

The 21st number is 11

The 2008th number is -1004 1005

The general formula is n (n+1).

So the 2007th number is 2007 2008

The general terms are not easy to find, but for the odd and even terms, you can use the Fibonacci sequence.

fn=1/√5[((1+√5)/2)^n-((1-√5)^2)]

a(2k-1)=f(k) f(k+1)(1 1,1 2,2 3,3 5,Cover the world's first ......）

a(2k)=f(k+1) f(k)(2 1,3 1,5 2,8 3,return to the mountain......）

First turn him into 1 1 2 4 6 9 10 16 15 25 · It is then 1 1 , 2 2 , 6 3 , 10 4 , 15 5 , and then (1) 1 , (1 + 1) 2 , (1 + 2 + 3) 3 , (1 + 2 + 3 + 4 ) 4 , (1 + 2 + 3 + 4 + 5) 5

Found the nth number (1+2+......n)/n²

The 98th number of a regular column of numbers 1, 1 3, -2 3, 3 3, 1 5, -2 5, 3 5, 4 5, -5 5, 1 7, 2 7, -3 7 7 is.

The 100th number is 19 19=1

The 99th number is -18 19

The 98th number is 17 19

3 = 3 to the power of 0 + 2

1 = 3 to the power of 1 -2

11 = 3 for the 2nd front cavity square +2

25 = 3 to the 3rd power -2

83 = 3 4 times Qingji Qingfang +2

3 to the 5th power - 2 = 241

So 3, 1, 11, 25, 83, (241).

Starting from the third number, each number is equal to the loss quotient of the first two short numbers +1