n(n+1)+(n+1)(n+2)+(n+2)(n+3).(n+2a)(n+2a+1)用一个代数式表示谢谢schumiandmassa 与guoshiqiangaxv的回答,
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n(n+1)+(n+1)(n+2)+(n+2)(n+3).(n+2a)(n+2a+1)用一个代数式表示谢谢schumiandmassa 与guoshiqiangaxv的回答,
n(n+1)+(n+1)(n+2)+(n+2)(n+3).(n+2a)(n+2a+1)
用一个代数式表示
谢谢schumiandmassa 与guoshiqiangaxv的回答,
n(n+1)+(n+1)(n+2)+(n+2)(n+3).(n+2a)(n+2a+1)用一个代数式表示谢谢schumiandmassa 与guoshiqiangaxv的回答,
(2a+1)*n^2+(2a+1)^2*n+4a(a+1)(2a+1)/3
这个符号不好打啊
2a
∑ (n+i)(n+i+1)
i=0
xz
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