an为等差数列 bn为等比数列a1=b1=2,a2-b2=1,a3=b3

1。A2=2+d A3=2+2dB2=2q B3=2q^22+d-2q=1 1+d=2q2+2d=2q^2 1+d=q^2解方程得q=2,d=3 q=0舍去,等比数列公比不为0An=3n-1Bn=q^n2。Tn=2/2^1+5/2^2+8/2^3+……+(3n-1)/2^n2Tn=2/2^0+5/2^1+8/2^2+……+(3n-1)/2^(n-1)2Tn-Tn=2+3[1/2^1+1/2^2+1/2^3+……1/2^(n-1)]-(3n-1)/2^n=2+3/2×(1-(1/2)^(n-1)/(1-(1/2))-(3n-1)/2^n=5-(3n+5)/2^nn>=1时3n+5>02^n>0(3n+5)/2^n>0所以Tn