Description
求[sum_{i = 1}^{n}i^m m^i , m leq 1000 ] 的值.
Solution
From Miskcoo's Space:
设 egin{eqnarray*} f(i) = sum_{k=1}^n k^i cdot m^k end{eqnarray*} 则我们要求$f(m)$.
所谓的"扰动法":
[egin{split}
(m-1)f(i) &= m cdot sum_{k = 1}^{n} k^i m^k - sum_{k = 1}^{n}k^i m^k \
&= sum_{i=1} ^ {n+1} (k-1)^i m^k - sum_{k = 1}^{n}k^i m^k \
&= n^i m^{n+1} + sum_{k = 1} ^n m^k sum_{j = 0}^{i-1} {i choose j} cdot (-1)^{i - j} cdot k^j \
&= n^i cdot m^{n + 1} + sum_{j = 0}^{i - 1} {i choose j} cdot (-1)^{i - j} sum_{k = 1}^n k^j cdot m^k \
&= n^i cdot m^{n + 1} + sum_{j = 0}^{i - 1} {i choose j} cdot (-1)^{i - j} cdot f(j) \
end{split}]
然后就变成了一个递推的问题.
基本思路是什么呢?从已知到未知,观察式子的特征进行转化来简化运算.但是自己什么时候才能把和式变换得这么溜呢....