1+(1+2)+...+(1+2+...+n)怎么化简啊?麻烦大家详细点,谢谢
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解决时间 2021-04-24 03:40
- 提问者网友:龅牙恐龙妹
- 2021-04-23 13:39
1+(1+2)+...+(1+2+...+n)怎么化简啊?麻烦大家详细点,谢谢
最佳答案
- 五星知识达人网友:躲不过心动
- 2021-04-23 13:57
S(n)=1+(1+2)+(1+2+3)+...+(1+2+...+n)
设an = 1+2+...+n = n(n+1)/2
则
Sn = a1 + a2 + ... + an
= 1(1+1)/2 + 2(2+1)/2 + ... + n(n+1)/2
= [1(1+1) + 2(2+1) + ... + n(n+1)]/2
= [(1*1 + 2*2 + ... + n*n) + (1 + 2 + ... + n)]/2
= [bn + an]/2
bn = 1*1 + 2*2 + ... + n*n
= n(n+1)(2n+1)/6
故
Sn = [bn + an]/2
= [n(n+1)(2n+1)/6 + n(n+1)/2]/2
设an = 1+2+...+n = n(n+1)/2
则
Sn = a1 + a2 + ... + an
= 1(1+1)/2 + 2(2+1)/2 + ... + n(n+1)/2
= [1(1+1) + 2(2+1) + ... + n(n+1)]/2
= [(1*1 + 2*2 + ... + n*n) + (1 + 2 + ... + n)]/2
= [bn + an]/2
bn = 1*1 + 2*2 + ... + n*n
= n(n+1)(2n+1)/6
故
Sn = [bn + an]/2
= [n(n+1)(2n+1)/6 + n(n+1)/2]/2
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