$$ 令\;0^0=1 \; ,\; f_x=\sum\limits^n_{i=0}i^x\times y^i。 \\ 则有\text{：}\\ \begin{align} f_x+(n+1)^x\times y^{n+1}\;&=\;\sum_{i=0}\limits^n(i+1)^n\times y^{i+1}+0\\ &=\sum_{i=0}\limits^n \bigg[\;y^{i+1}\big( \sum_{j=0}\limits^xC^j_x\times i^j \big) \bigg]\\ &=\sum_{i=0}\limits^n\sum_{j=0}\limits^xy^{i+1}i^jC^j_x\\ &=y\times\; \sum_{j=0}\limits^x \bigg(\; \sum_{i=0}\limits^ni^j\times y^i \; \bigg)\\ &=y\times\; \sum_{j=0}\limits^x C^j_xf_j \end{align} \\ \therefore f_j=\; \bigg(\;\; (n+1)^x\times y^{n+1}-y\times \sum_{j=0}\limits^x C^j_xf_j \;\;\bigg)\;\times \;(y-1)^{-1}\;\;\;[y\not =1] $$

$$ 此时，若\;y==1\;,则有 \text ：\\ \begin{align} (n+x)^x\times y^{n+1}\;&=\;y\times \sum_{j=0}\limits^{x-2} C^j_xf_j\,+\,y\times C^{x-1}_x\times f_{x-1}\\ &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 令\;x=x+1\;，得\text：\\ (n+x)^{x+1}\times y^{n+1}\;&=\;y\times \sum_{j=0}\limits^{x-1} C^j_xf_j\,+\,y\times C^{x}_{x+1}\times f_x\\\\ &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 进而得\text ：\\ f_x\;&=\;\bigg[\;\; (n+x)^{x+1}\times y^{n+1}\,-\,\sum_{j=0}\limits^{x-1} y\,C^j_xf_j \;\;\bigg]\,\times [\, y\times C^x_{x+1} \,]^{-1} \end{align} $$