前14个自然数幂求和公式

\[\begin{aligned} \sum_{k=1}^{n}k^{0}&=n \\\sum_{k=1}^{n}k^{1}&=\frac12n^2+\frac12n \\\sum_{k=1}^{n}k^{2}&=\frac13n^3+\frac12n^2+\frac16n \\\sum_{k=1}^{n}k^{3}&=\frac14n^4+\frac12n^3+\frac14n^2 \\\sum_{k=1}^{n}k^{4}&=\frac15n^5+\frac12n^4+\frac13n^3-\frac1{30}n \\\sum_{k=1}^{n}k^{5}&=\frac16n^6+\frac12n^5+\frac5{12}n^4-\frac1{12}n^2 \\\sum_{k=1}^{n}k^{6}&=\frac17n^7+\frac12n^6+\frac12n^5-\frac16n^3+\frac1{42}n \\\sum_{k=1}^{n}k^{7}&=\frac18n^7+\frac12n^7+\frac7{12}n^6-\frac7{24}n^4+\frac1{12}n^2 \\\sum_{k=1}^{n}k^{8}&=\frac19n^7+\frac12n^8+\frac23n^7-\frac7{15}n^5 +\frac29n^3-\frac1{30}n \\\sum_{k=1}^{n}k^{9}&=\frac1{10}n^{10}+\frac12n^9+\frac34n^8-\frac7{10}n^6+\frac12n^4-\frac3{20}n^2 \\\sum_{k=1}^{n}k^{10}&=\frac1{11}n^{11}+\frac12n^{10}+\frac56n^9-n^7+n^5-\frac12n^3+\frac5{66}n \\\sum_{k=1}^{n}k^{11}&=\frac1{12}n^{12}+\frac12n^{11}+\frac{11}{12}n^{10}-\frac{11}8n^8+\frac{11}6n^6-\frac{11}8n^4+\frac5{12}n^2 \\\sum_{k=1}^{n}k^{12}&=\frac1{13}n^{13}+\frac12n^{12}+n^{11}-\frac{11}6n^9+\frac{22}7n^7-\frac{33}{10}n^5+\frac53n^3-\frac{691}{2730}n \\\sum_{k=1}^{n}k^{13}&=\frac1{14}n^{14}+\frac12n^{13}+\frac{13}{12}n^{12}-\frac{143}{60}n^{10}+\frac{143}{28}n^8-\frac{143}{20}n^6+\frac{65}{12}n^4-\frac{691}{420}n^2 \end{aligned} \]