| 2] 其规律是 Sum(k=0,n)[(2k+1)^2] = |
| 送交者: zhf 2019月02月24日07:44:18 于 [灵机一动] 发送悄悄话 |
| 回 答: 趣味的数学 - 13【数字模式】 由 gugeren 于 2019-02-22 21:15:45 |
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其规律是 Sum(k=0,n)[(2k+1)^2] = (2n+1)(2n+2)(2n+3)/6 用数学归纳法证明。 n=0时明显成立。 Sum(k=0,n+1)[(2k+1)^2] = (2n+1)(2n+2)(2n+3)/6 + (2(n+1)+1)^2 = (2(n+1)+1)[(2n+1)(2n+2)/6 + 2(n+1)+1] = (2(n+1)+1)[(2n+1)(2n+2)+ 12(n+1)+6]/6 = (2(n+1)+1)[(4n^2 + 18n + 20]/6 = (2(n+1)+1)[(2n+4)(2n+5)]/6 = (2(n+1)+1)[(2(n+1)+2)(2(n+1)+3)]/6 |
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