[√x+(1/2)×(1/x)^(1/4)]^n 展开式通项T(r+1)=C(n,r)[(√x)^(n-r)][(1/2)×(1/x)^(1/4)]^r=C(n,r)[(1/2)^r][(√x)^(n-r)][(1/x)^(r/4)],取r=0,1,2得前三项系数C(n,0)[(1/2)^0,C(n,1)[(1/2)^1,C(n,2)[(1/2)^2,即1,n/2,n(n-1)/8,成等差数列,1+n(n-1)/8=n,n2-9n+8=0,n>1,n=8,[√x+(1/2)×(1/x)^(1/4)]^8 展开式通项T(r+1)=C(8,r)[(1/2)^r][(√x)^(8-r)][(1/x)^(r/4)]=C(8,r)[...
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