Question

1的k次方+2的k次方+3的k次方+…+n的k次方=?

帮忙写出求和的方法，谢谢指导
是否有与杨辉三角有关的方法

Answer 1

偶只能弄出递推公式，具体的解析式太复杂了。。。。
考虑(n+1)^k-n^k=Ck 1 *n^(k-1)+Ck 2 *n^(k-2)+...+Ck (k-1) *n^1 +1
n^k-(n-1)^k=Ck 1 *(n-1)^(k-1)+...+Ck (k-1) *(n-1)^1 +1
...
2^k-1^k =Ck 1*1^(k-1)+...+Ck (k-1) *1^1 +1
令1^k+..+n^k=S[n,k],上述叠加，得
（n+1)^k-1=Ck 1 *S[n,k-1] + Ck 2 *S[n,k-2]+...+Ck (k-1) *S[n,1]+n
n^k -1=Ck 1 *S[n-1,k-1]+...+Ck (k-1) *S[n-1,1]+n
...
2^k-1=Ck 1 *S[1,k-1]+...+Ck (k-1) *S[1,1]+n
叠加得 (n+1)^k+...+2^k-n=Ck 1 *{S[n,k-1]+...+S[1,k-1]}+...+n^n

∴n^k+...+2^k+1^k=S[n,k]=Ck 1 *{S[n,k-1]+...+S[1,k-1]}+...+n^n+n+1-(n+1)^k