设数列an的前n项和为sn,点P(Sn,an)在直线(3-m)x+2my-m-3=0上,m属于N*,m为常数,m≠-3.(1)求an(2)若数列an的公比q=f(m),数列bn满足b1=a1,bn=3/2*f[b(n-1)],n属于N*,n≥2,求证{1/bn}为等差数列,并求bn(3)设数列cn满足cn=bn

Question

设数列an的前n项和为sn,点P(Sn,an)在直线(3-m)x+2my-m-3=0上,m属于N*,m为常数,m≠-3.
(1)求an
(2)若数列an的公比q=f(m),数列bn满足b1=a1,bn=3/2*f[b(n-1)],n属于N*,n≥2,求证{1/bn}为等差数列,并求bn
(3)设数列cn满足cn=bn*b(n+2),Tn为数列cn的前n项和,且存在实数T满足Tn≥T,(n属于N*),求T的最大值

Answer 1

(1)p点代入直线方程：（3-m）Sn+2m*an-m-3=0;：（3-m）S(n-1）+2m*a（n-1）-m-3=0.相减得（3+m）an=2m*a（n-1）,an/a(n-1）=2m/(3+m),an为等比数列,a1=1,an=[2m/(3+m)]^(n-1);
(2)f(m)=2m/(3+m).bn=bn=3/2*f[b(n-1)]=3/2*[2b(n-1)/(3+b(n-1))]=3b(n-1)/(3+b(n-1)).1/bn=1/b(n-1)+1/3,所以,{1/bn}为等差数列,1/bn=1/b1+(n-1)d=1+(n-1)=n;
(3)cn=bn*b(n+2)=n*(n+2)=n^2+2n,Tn为数列cn的前n项和,Tn=n(n+1)(2n+1)/6+2*(1+n)n/2=n(n+1)(2n+7)/6,因为n属于N*,当n取1时Tn值最小,Tn（min）=3；则t