已知数列{an}是等差数列.a3=10,a6=22数列{bn}的前n项和为Sn且Sn+1/3bn=1求{an}的通项公式证明{bn}是等比数列设数列{an}的前n项和为Sn且2an=Sn+2n+1n属于n*求a1a2a3求证数列{an+2}是等比数列求{n*an}的前n项

Question

已知数列{an}是等差数列.a3=10,a6=22数列{bn}的前n项和为Sn且Sn+1/3bn=1求{an}的通项公式 证明{bn}是等比
数列
设数列{an}的前n项和为Sn且2an=Sn+2n+1 n属于n*求a1 a2 a3求证数列{an+2}是等比数列求{n*an}的前n项和Tn

Answer 1

a6=a3+3d
22=10+3d
3d=12
d=4
an=a1+(n-1)d
=a3+(n-3)d
=10+4(n-3)
=4n-2
sn+1/3bn=1
s(n-1)+1/3b(n-1)=1
两式相减得：
sn+1/3bn-s(n-1)-1/3b(n-1)=0
bn+1/3bn-1/3b(n-1)=0
4/3bn-1/3b(n-1)=0
4bn-b(n-1)=0
bn/b(n-1)=1/4
所以bn为等比数列.
2a1=a1+2*1+1
2a1=a1+3
a1=3
2a2=S2+2*2+1
2a2=a1+a2+5
a2=a1+5
a2=3+5=8
2a3=S3+2*3+1
2a3=a1+a2+a3+7
a3=a1+a2+7
a3=3+8+7=18
2an=Sn+2n+1
Sn=2an-2n-1
S(n-1)=2a(n-1)-2(n-1)-1=2a(n-1)-2n+1
an=Sn-S(n-1)
an=2an-2n-1-2a(n-1)+2n-1
an=2an-2a(n-1)-2
an=2a(n-1)+2
an+2=2a(n-1)+4
[an+2]/[a(n-1)+2]=2
所以{an+2}是以2为公比的等比数列
an+2=(a1+2)q^(n-1)
an+2=(3+2)*2^(n-1)
an+2=5*2^(n-1)
an=5*2^(n-1)-2
Tn=1*[5*2^(1-1)-2]+2*[5*2^(2-1)-2]+.+n*[5*2^(n-1)-2]
=5*1*2^0-2*1+5*2*2^1-2*2+.+5n*2^(n-1)-2n
=5*1*2^0+5*2*2^1+.+5n*2^(n-1)-2*1-2*2-.-2n
=5[1*2^0+2*2^1+.+n*2^(n-1)]-2(1+2+.+n)
=5[1*2^0+2*2^1+.+n*2^(n-1)]-n(n+1)
2Tn=5[1*2^1+2*2^2+.+n*2^n]-2n(n+1)
Tn-2Tn=5*[2^0+2^1+2^2+.+2^(n-1)]-5n*2^n-n(n+1)+2n(n+1)
Tn-2Tn=5*[1*(1-2^n)/(1-2)]-5n*2^n+n(n+1)
-Tn=5*(2^n-1)-5n*2^n+n(n+1)
Tn=5n*2^n-5*(2^n-1)-n(n+1)
Tn=5n*2^n-5*2^n+5-n(n+1)
Tn=5(n-1)*2^n-n(n+1)+5