作者suhorng ( )
站内Math
标题Re: [中学] 证1*2+2*3+3*4+...+n*(n+1)=n(n+1)(n+2)/3
时间Mon Jun 20 18:36:02 2011
我觉得不用展开. 观察
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(n+1)(n+2)(n+3)...(n+k+1)-n(n+1)(n+2)...(n+k) = (k+1)*(n+1)(n+2)...(n+k)
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设 a_n = (n+1)(n+2)...(n+k), b_n = n(n+1)(n+2)...(n+k)
有 b_{n+1} - b_n = (k+1)a_n
所以 a_0 + a_1 + ... + a_{n-1} = (b_n - b_0)/(k+1) = n(n+1)(n+2)...(n+k)/(k+1)
※ 引述《MinusHeart (ICRT)》之铭言:
: 证1*2+2*3+3*4+...+n*(n+1)=n(n+1)(n+2)/3
: 在做离散数学中的题目
: 1*2+2*3+3*4+...+n*(n+1)=
: 2+6+12+20+30+....+n*(n+1)=
: 证到这就卡住了
: 想请各位高手帮忙解答
: 谢谢!!
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※ 编辑: suhorng 来自: 59.115.147.134 (06/20 18:36)
1F:推 MinusHeart :感谢指教@@ 06/20 20:19