帮忙解决四道数列规律题~(1)56,66,78,82,(?)(2)9,1,4,3,40,(?)(3)20,22,25,30,37,(?)(4)124,3612,51020,(?)要详细的规律解释,谢谢,答案分别是:98,121,48,71428
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 04:05:48
![帮忙解决四道数列规律题~(1)56,66,78,82,(?)(2)9,1,4,3,40,(?)(3)20,22,25,30,37,(?)(4)124,3612,51020,(?)要详细的规律解释,谢谢,答案分别是:98,121,48,71428](/uploads/image/z/1997732-20-2.jpg?t=%E5%B8%AE%E5%BF%99%E8%A7%A3%E5%86%B3%E5%9B%9B%E9%81%93%E6%95%B0%E5%88%97%E8%A7%84%E5%BE%8B%E9%A2%98%7E%EF%BC%881%EF%BC%8956%2C66%2C78%2C82%2C%EF%BC%88%3F%EF%BC%89%EF%BC%882%EF%BC%899%2C1%2C4%2C3%2C40%2C%EF%BC%88%3F%EF%BC%89%EF%BC%883%EF%BC%8920%2C22%2C25%2C30%2C37%2C%EF%BC%88%3F%EF%BC%89%EF%BC%884%EF%BC%89124%2C3612%2C51020%2C%EF%BC%88%3F%EF%BC%89%E8%A6%81%E8%AF%A6%E7%BB%86%E7%9A%84%E8%A7%84%E5%BE%8B%E8%A7%A3%E9%87%8A%EF%BC%8C%E8%B0%A2%E8%B0%A2%EF%BC%8C%E7%AD%94%E6%A1%88%E5%88%86%E5%88%AB%E6%98%AF%EF%BC%9A98%EF%BC%8C121%EF%BC%8C48%EF%BC%8C71428)
帮忙解决四道数列规律题~(1)56,66,78,82,(?)(2)9,1,4,3,40,(?)(3)20,22,25,30,37,(?)(4)124,3612,51020,(?)要详细的规律解释,谢谢,答案分别是:98,121,48,71428
帮忙解决四道数列规律题~
(1)56,66,78,82,(?)
(2)9,1,4,3,40,(?)
(3)20,22,25,30,37,(?)
(4)124,3612,51020,(?)
要详细的规律解释,谢谢,答案分别是:98,121,48,71428
帮忙解决四道数列规律题~(1)56,66,78,82,(?)(2)9,1,4,3,40,(?)(3)20,22,25,30,37,(?)(4)124,3612,51020,(?)要详细的规律解释,谢谢,答案分别是:98,121,48,71428
56-(5+6)=45=5*9
66-(6+6)=54=6*9
78-(7+8)=63=7*9
82-(8+2)=72=8*9
因为:98-(9+8)=81=9*9
所以答案是:98
把每一个都除以3,看余数
9÷3余数为0;
1÷3余数为1;
4÷3余数为1;
3÷3余数为0;
40÷3余数为1;
由两个余数为1的数,后都减1÷前都有一个倍数,即:n=(4-1)÷1=3;则有(x-1)÷40=3(x为第六个数)
即为:x=40×3+1=121
质数相加,22+2=22
22+3=25
25+5=30
那个空应是48.
第一项*2=第二项 第一项*4=第三项 (2n-1)((2n-1)*2)((2n-1)*4)
124=1(1*2)(1*4)=124
3612=3(3*2)(3*4)
51020=5(5*2)(5*4)
71428=7(7*2)(7*4)
(1)98 (3)48(4)71428(2)不会