数列 -1,0,27 ( )

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/05/15 01:07:27

数列 -1,0,27 ( )
答案为128

这题其实可以根据不同的规律找到不同的答案,并没有唯一正确的答案
-1 = -1×1² = (0²-1)×1²
0 = 0×2² = (1²-1)×2²
27 = 3×3² = (2²-1)×3²
128 = 8×4² = (3²-1)×4²
375 = 15×5² = (4²-1)×5²
……
a(n) = [(n-1)²-1]×n²=n^4-2×n³
即是题中答案给出的解
-1 = (-1)³ = (0²-1)³
0 = 0³ = (1²-1)³
27 = 3³ = (2²-1)³
512 = 8³ = (3²-1)³
3375 = 15³ = (4²-1)³
……
a(n) = [(n-1)²-1]³
即是楼上给出的解
还可以
-1 = -1×1^1 = (1-2)×1^1
0 = 0×2^2 = (2-2)×2^2
27 = 1×3^3 = (3-2)×3^3
512 = 2×4^4 = (4-2)×4^4
9375 = 3×5^5 = (5-2)×5^5
……
a(n) = (n-2)×n^n
-1 = -1×1^5 = (1-2)×1^(6-1)
0 = 0×2^4 = (2-2)×2^(6-2)
27 = 1×3^3 = (3-2)×3^(6-3)
32 = 2×4^2 = (4-2)×4^(6-4)
15 = 3×5^1 = (5-2)×5^(6-5)
……
a(n) = (n-2)×n^(6-n)
-1 = -1×3^(0/2) = (1-2)×3^[3(1-1)/2]
0 = 0×3^(3/2) = (2-2)×3^[3(2-1)/2]
27 = 1×3^(6/2) = (3-2)×3^[3(3-1)/2]
162√3 = 2×3^(9/2) = (4-2)×3^[3(4-1)/2]
2187 = 3×3^(12/2) = (5-2)×3^[3(5-1)/2]
……
a(n) = (n-2)×3^[3(n-1)/2]

数列1(数列) 数列1 数列1 1 6 5 20 27 一个数列.1 5 16 27 ( ). 数列：—1,0,1,（）64, 数列-1,0,1,8,(),64, 数列1,4,9,( ),1,0 不等式证明与数列1,若0 0,1,3,6,10是什么数列已知数列的前四项为1,3,9,27,求数列的通项公式求数列的通项公式,数列：0,1,3,6,10,15,21… 数列-1，85