作业帮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(
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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(
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
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(
(1)Sn^2=S(n-1)^2+1=S(n-2)^2+1+1=S(n-3)^2+3=.=S1^2+(n-1)=n+2
Sn=√(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-5
bn=(4n-5)*2^n
Tn=4*(1*2+2*2^2+3*2^3+.+n*2^n)-5(2+2^2+2^3+...+2^n)
设M=1*2+2*2^2+3*2^3+.+n*2^n
2M= 1*2^2+2^2^3+.+(n-1)*2^n+n*2^(n+1)
M=-2-2^2-2^3-.-2^n+n*2^(n+1)
Tn=-9(2+2^2+2^3+...+2^n)+4*n*2^(n+1)
=(4n-9)*2^(n+1)+18
(3)因为lgx+lgy=1,所以xy=10
Sn=lg[x^n*x^(n-1)y*x^(n-2)y^2*.*xy(n-1)*y^n]
=lg{x^[n+(n-1)+(n-2)+.+2+1]*y^[1+2+...+(n-1)+n]}
=lg(xy)^[n(n+1)/2]
=n(n+1)/2