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第二項的二項式係數為nC1第三項的二項式係數為nC2第四項的二項式係數為nC3因為成等差數列所以nC1+nC3=2*nC2n+n*(n-1)*(n-2)/6=2*n*(n-1)/21+(n^2-3n+2)/6=n-16+n^2-3n+2=6n-6n^2-9n+14=0(n-2)*(n-7)=0所以n=2(舍)…
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