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已知数列an=n²,求数列的前n项和Sn.
人气:206 ℃ 时间:2020-05-27 09:59:24
解答
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)(2n+1)
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