作业帮1的k次方+2的k次方+3的k次方+4的k次方+……(n-1)的k次方+n的k次方=?
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/27 18:44:50
1的k次方+2的k次方+3的k次方+4的k次方+……(n-1)的k次方+n的k次方=?
1的k次方+2的k次方+3的k次方+4的k次方+……(n-1)的k次方+n的k次方=?
1的k次方+2的k次方+3的k次方+4的k次方+……(n-1)的k次方+n的k次方=?
s(k)=1^k + 2^k + ... + (n-1)^k + n^k,
(1+x)^(k+1) = 1 + c(k+1,1)x^1 + c(k+1,2)x^2 + ... + c(k+1,k)x^k + x^(k+1),
(1+x)^(k+1) - x^(k+1) = 1 + c(k+1,1)x^1 + c(k+1,2)x^2 + ... + c(k+1,m)x^m + ... + c(k+1,k)x^k,
让x分别为1,2,., (n-1),n.
(1+1)^(k+1)-1^(k+1) = 1 + c(k+1,1)1^1 + c(k+1,2)1^2 + ... + c(k+1,m)1^m + ... + c(k+1,k)1^k,
(2+1)^(k+1)-2^(k+1) = 1 + c(k+1,1)2^1 + c(k+1,2)2^2 + ... + c(k+1,m)2^m + ... + c(k+1,k)2^k,
...
(n-1+1)^(k+1) - (n-1)^(k+1) = 1 + c(k+1,1)(n-1)^1 + c(k+1,2)(n-1)^2 + ... + c(k+1,m)(n-1)^m + ... + c(k+1,k)(n-1)^k,
(n+1)^(k+1) - n^(k+1) = 1 + c(k+1,1)n^1 + c(k+1,2)n^2 + ... + c(k+1,m)n^m + ... + c(k+1,k)n^k,
上面n个等式,等号两边分别求和,有,
(n+1)^(k+1) - 1^(k+1) = n + c(k+1,1)s(1) + c(k+1,2)s(2) + ... + c(k+1,m)s(m) + ... + c(k+1,k)s(k).
.（*）
我们知,
c(k+1,m) = (k+1)!/[m!(k+1-m)!], m=1,2,...,k,
s(1) = 1 + 2 + ... + n = n(n+1)/2,
s(2) = 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6,
s(3) = 1^3 + 2^3 + ... + n^3 = [n(n+1)/2]^2,
将这些已知的s(1),s(2),s(3)带入（*）,并令k=4,可得 s(4).
将s(1),s(2),s(3),s(4)带入(*), 并令k=5,可得s(5),
...
将s(1),s(2),...,s(p-1) 带入(*),并令 k=p,可得s(p).
建议你用假设法