设数列an的前n项和为Sn,且2an=Sn+2n+1 求a1 a2 a3 求证:数列{an+2}是等比数列 ...设数列an的前n项和为Sn,且2an=Sn+2n+1求a1 a2 a3求证:数列{an+2}是等比数列求数列{n*an}的前n项和Tn

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设数列an的前n项和为Sn,且2an=Sn+2n+1 求a1 a2 a3 求证:数列{an+2}是等比数列 ...设数列an的前n项和为Sn,且2an=Sn+2n+1求a1 a2 a3求证:数列{an+2}是等比数列求数列{n*an}的前n项和Tn
设数列an的前n项和为Sn,且2an=Sn+2n+1 求a1 a2 a3 求证:数列{an+2}是等比数列 ...
设数列an的前n项和为Sn,且2an=Sn+2n+1
求a1 a2 a3
求证:数列{an+2}是等比数列
求数列{n*an}的前n项和Tn

设数列an的前n项和为Sn,且2an=Sn+2n+1 求a1 a2 a3 求证:数列{an+2}是等比数列 ...设数列an的前n项和为Sn,且2an=Sn+2n+1求a1 a2 a3求证:数列{an+2}是等比数列求数列{n*an}的前n项和Tn
由2an=Sn+2n+1
2a(n-1)=S(n-1)+2(n-1)+1,得：
an-2a(n-1)=2 （n>1）…… ①
1.令n=1,可得：a1=3.
带入①式,易得：a2=9,a3=20；
2.①式即为an+2=2[a(n-1)+2],故有：
(an+2)/[a(n-1)+2] =2,所以an+2是公比为2的等比数列.
3.由上式可求的n*an的通项,再求Tn即可

2an=Sn+2n+1
2a1=a1+2+1
a1=3 a2=8 a3=18
2an=Sn+2n+1 (1)
2a(n-1)=S(n-1)+2(n-1)+1 (2)
(1)-(2)
an-2a(n-1)=2 ...

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2an=Sn+2n+1
2a1=a1+2+1
a1=3 a2=8 a3=18
2an=Sn+2n+1 (1)
2a(n-1)=S(n-1)+2(n-1)+1 (2)
(1)-(2)
an-2a(n-1)=2
an+2=2[a(n-1)+2]
(an+2)/[a(n-1)+2] =2
数列{an+2}等比；公比为2；首项5
an+2=5*2^(n-1)
an+2=5*2^(n-1)-2
n*an=5n2^(n-1)-2n (分两项求和）
Tn1=5*[1+2*2^1+3*2^2+……+n2^(n-1)] (1)
2Tn1= 5*[ 2 +2*2^3+3*2^3+……+n2^n] (2)
(1)-(2)得
Tn1-2Tn1=-Tn=5[1+2^1+2^3+……+2^(n-1)-n2^n]
=5[ (2^n-1)/(2-1) -n2^n]
Tn1= 5[n2^n-2^n+1 ]
Tn2=2(1+2+3……+n)=n(n+1)
Tn=Tn1- Tn2
= 5[n2^n-2^n+1 ] -n(n+1)
=ok

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