There is a set of numbers 1 2 2 4 8 32 Find the hundredth term of this series

public class test{

public static void main (string args){

long a = new long[100];

a[0] = 1;

a[1] = 2;

for(int i= 2;i<100;i++)a[i] = a[i-1]*a[i-2];

Section"+(i+1)+"Number:"+a[i]);

It can only be counted to the 11th digit, 1, 2, 2, 4, 8, 32, 256, 8192, 2097152, 17179869184, 36028797018963968

You three three look, 121, 232, 343, 454,So 99-101 is 33 34 35, then the 100th should be 34

According to the title, it can be seen that this number column is the first term A1 2, and the tolerance is D 3.

So an a1+(n-1)d 2+3n-3 3n-1, so sn (a1+an)n 2=(2+3n-1)n 2=(3n+1)n 2

So s100 (3 100+1) 100 2 15050

This is a first term that is 2,2+(100-1) 3

Answer: The 100th late number in this column is 299

Therefore, the answer is: 299

3 1 set. 100 3 = 33 surplus 1

The 34th group is 34, 35, 36

So the 100th is 34

This is a series of equal differences, a1=6, d=4, an=a1+(n-1)da100=a1+(100-1)d=6+99 4=402, the first n terms are the sum of sn=(a1+an)n 2

The sum of the first 100 items is s100 = (a1 + a100) 100 2 = (6 + 406) 100 2 = 20400.

20400, the child goes back to memorize more formulas, equal difference series, proportional number series formulas, these are basic.