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先看他們的絕對值 bn=|an| 則 b2-b1=4 b3-b2=6 b4-b3=8 …… bn-b(n-1)=2n 相加 bn-b1=2*(2+3+……+n)=2*(n+2)(n-1)/2=n^2+n-2 b1=2 bn=n^2+n 所以an=(-1)^(n+1)*(n^2+n)
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