{√x+1/[2x^(1/4)]}^n的展开式中,
T=C(n,r)(√x)^(n-r)*[(1/2)x^(-1/4)]^r
=C(n,r)*(1/2)^r*x^(n/2-3r/4),
(1)前三项系数成等差数列,
∴1+C(n,2)/4=n,1+n(n-1)/8=n,n^2-9n+8=0,n>1,∴n=8.
(2)第三项的二项式系数=C(8,2)=8*7/2=28,项的系数=C(8,2)/4=7.
(3)4-3r/4=1,r=4,
T5的系数=C(8,4)/2^4=35/8,为所求.