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常见自然数列和

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求和 展开式 求和公式
\(\sum_{k=1}^n k\) = 1 + 2 + 3 + 4 + .. + n = (n2 + n) / 2
= (1/2)n2 + (1/2)n
\(\sum_{k=1}^n k^2\) = 1 + 4 + 9 + 16 + .. + n2 = (1/6)n(n+1)(2n+1)
= (1/3)n3 + (1/2)n2 + (1/6)n
\(\sum_{k=1}^n k^3\) = 1 + 8 + 27 + 64 + .. + n3 = (1/4)n4 + (1/2)n3 + (1/4)n2
\(\sum_{k=1}^n k^4\) = 1 + 16 + 81 + 256 + .. + n4 = (1/5)n5 + (1/2)n4 + (1/3)n3 - (1/30)n
\(\sum_{k=1}^n k^5\) = 1 + 32 + 243 + 1024 + .. + n5 = (1/6)n6 + (1/2)n5 + (5/12)n4 - (1/12)n2
\(\sum_{k=1}^n k^6\) = 1 + 64 + 729 + 4096 + .. + n6 = (1/7)n7 + (1/2)n6 + (1/2)n5 - (1/6)n3 + (1/42)n
\(\sum_{k=1}^n k^7\) = 1 + 128 + 2187 + 16384 + .. + n7 = (1/8)n8 + (1/2)n7 + (7/12)n6 - (7/24)n4 + (1/12)n2
\(\sum_{k=1}^n k^8\) = 1 + 256 + 6561 + 65536 + .. + n8 = (1/9)n9 + (1/2)n8 + (2/3)n7 - (7/15)n5 + (2/9)n3 - (1/30)n
\(\sum_{k=1}^n k^9\) = 1 + 512 + 19683 + 262144 + .. + n9 = (1/10)n10 + (1/2)n9 + (3/4)n8 - (7/10)n6 + (1/2)n4 - (3/20)n2
\(\sum_{k=1}^n k^{10}\) = 1 + 1024 + 59049 + 1048576 + .. + n10 = (1/11)n11 + (1/2)n10 + (5/6)n9 - n7 + n5 - (1/2)n3 + (5/66)n