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1+(n+2)+(2n+3)+(3n+4)+(4n+5)+……((n-1)n+n)的答案
人气:181 ℃ 时间:2020-03-27 12:08:35
解答
1+(n+2)+(2n+3)+(3n+4)+(4n+5)+……((n-1)n+n)
=(1+2+3+...+n)+(n+2n+3n+.(n-1)n)
=(1+n)n/2+n(1+2+3+.(n-1))
=(1+n)n/2+n*(1+n-1)(n-1)/2
=(1+n)n/2+n^2(n-1)/2
=n/2(1+n+n^2-n)
=n(n^2+1)/2
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