1.设Sn是正数数列{an}的前n项和,已知数列S1 ^2,S2 ^2,...Sn ^2...是以3为首项,1为公差的等差数列.求_数学解答

1.设Sn是正数数列{an}的前n项和,已知数列S1 ^2,S2 ^2,...Sn ^2...是以3为首项,1为公差的等差数列.求{an}通项公式
2.已知数列{an}前n项和Sn=2n^2-3n,若bn=an·2^n,求数列{bn}的前n项和Tn.
3.已知lgx+lgy=1,Sn=lg x^n+lg(x^(n-1)y)+lg(x^(n-2)y^2).+lg(xy^(n-1))+lgy^n,求Sn

人气：199 ℃　时间：2020-06-14 18:56:02

解答

(1)Sn^2=S(n-1)^2+1=S(n-2)^2+1+1=S(n-3)^2+3=.=S1^2+(n-1)=n+2Sn=√(n+2) an=Sn-S(n-1)=√(n+2)-√(n+1) (当n≥2时成立,a1=√3）(2)an=Sn-S(n-1)=2n^2-3n-2(n-1)^2+3(n-1)=4n-5bn=(4n-5)*2^nTn=4*(1*2+2*2^2+3*2^3+...