n²的前n项和怎么求
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解决时间 2021-03-29 00:42
- 提问者网友：疯孩纸
- 2021-03-28 09:47
n²的前n项和怎么求
最佳答案
- 五星知识达人网友：不想翻身的咸鱼
- 2021-03-28 10:10
an
= n^2
= n(n+1) -n
=(1/3)[ n(n+1)(n+2) -(n-1)n(n+1) ] - (1/2)[n(n+1) -(n-1) n]
Sn
=a1+a2+...+an
=(1/3)n(n+1)(n+2) - (1/2)n(n+1)
=(1/6)n(n+1)[ 2(n+2) -3]
=(1/6)n(n+1)( 2n+1)
= n^2
= n(n+1) -n
=(1/3)[ n(n+1)(n+2) -(n-1)n(n+1) ] - (1/2)[n(n+1) -(n-1) n]
Sn
=a1+a2+...+an
=(1/3)n(n+1)(n+2) - (1/2)n(n+1)
=(1/6)n(n+1)[ 2(n+2) -3]
=(1/6)n(n+1)( 2n+1)
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