Question

设数列{an}的前n项和Sn满足2Sn=n2+n,{bn}为等比数列,且a1=b1,b2=2(a2-a1)(1)求an,bn（2）

设数列{an}的前n项和Sn满足2Sn=n2+n,{bn}为等比数列,且a1=b1,b2=2(a2-a1)(1)求an,bn（2）cn=anbn,求数列｛cn｝的前n项和Tn

Answer 1

解：(1)由2Sn=n^2+n得Sn=(n^2+n)/2
a1=S1=(1^2+1)/2=1
b1=a1=1
a2+a1=S2=(2^2+2)/2=3
a2=3-a1=2
b2=2(a2-a1)=2
等比数列{bn}的公比为q=b2/b1=2
bn=b1q^(n-1)=2^(n-1)
S(n-1)=[(n-1)^2+n-1]/2
an=Sn-S(n-1)=(n^2+n)/2-[(n-1)^2+n-1]/2=n
所以an=n；bn=2^(n-1)
(2)cn=anbn=n2^(n-1)
设等比数列前n项的和为Bn
则Tn=b1+2B2+3B3+4B4+...+nBn
=Bn+Bn-B1+Bn-B2+Bn-B3+Bn-B4+...+Bn-B(n-1)
=nBn-[B1+B2+B3+B4+...B(n-1)]
=nb1(1-q^n)/(1-q)-{b1(1-q)/(1-q)+b1(1-q^2)/(1-q)+b1(1-q^3)/(1-q)...+b1[1-q^(n-1)]/(1-q)}
=n(1-2^n)/(1-2)-{(1-2)/1-2)+(1-2^2)/(1-2)+(1-2^3)/(1-2)....[1-2^(n-1)]/(1-2)}
=n(2^n-1)-{(2-1)+(2^2-1)+(2^3-1)+....+[2^(n-1)-1]}
=n2^n-n-[2+2^2+2^3+....2^(n-1)-(n-1)]
=n2^n-n-{2[1-2^(n-1)]/(1-2)-(n-1)}
=n2^n-n-[(2-2^n)/(-1)-n+1]
=n2^n-n-(2^n-2-n+1)
=n2^n-n-2^n+1+n
=(n-1)2^n+1

Answer 2

（1）
Sn=n(n+1)/2
an=n为等差数列
所以b1=1，b2=2(2-1)=2，q=2/1=2
所以bn=2^(n-1)
（2）
令Bn为bn前n项和，则Bn=2^n-1
Tn=Sn+Bn=n(n+1)/2+2^n-1