数列An满足:A1=1,A2=2/3,An+2=3/2An+1-1/2An,记dn=An+1-An,求证dn是等比数列(2)求数列An的通项公式(3)令Bn=3n-2,求数列An*Bn的前n项和Sn
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数列An满足:A1=1,A2=2/3,An+2=3/2An+1-1/2An,记dn=An+1-An,求证dn是等比数列(2)求数列An的通项公式(3)令Bn=3n-2,求数列An*Bn的前n项和Sn
数列An满足:A1=1,A2=2/3,An+2=3/2An+1-1/2An,记dn=An+1-An,求证dn是等比数列
(2)求数列An的通项公式
(3)令Bn=3n-2,求数列An*Bn的前n项和Sn
数列An满足:A1=1,A2=2/3,An+2=3/2An+1-1/2An,记dn=An+1-An,求证dn是等比数列(2)求数列An的通项公式(3)令Bn=3n-2,求数列An*Bn的前n项和Sn
(1)a(n+2)=3/2a(n+1)-1/2 an
a(n+2)-a(n+1)=1/2 a(n+1)-1/2 an
d(n+1)=(1/2)* dn
d(n+1)/dn =1/2
所以{dn}为等比数列 ,q=1/2 首项为 1/2
(2)a(n+1)-an=(1/2)^n
an-a(n-1)=(1/2)^(n-1) ,n≥2
...
a2-a1=1/2
累加,得 an-a1=1/2+(1/2)^2+...+(1/2)^(n-1)
=1-(1/2)^(n-1)
an=1-(1/2)^(n-1)+a1= 2 -(1/2)^(n-1)
a1= 2-(1/2)^0=2-1=1
所以an=2 -(1/2)^(n-1) ,n∈N*
(3) bn=3n-2
an*bn=【2 -(1/2)^(n-1)】*(3n-2)=6n-4 - (3n-2) (1/2)^(n-1)
Sn= (2+8+14+,+6n-4) - (1*(1/2)^0+4*(1/2)^1 + 7 * (1/2)^2 +...+(3n-2) (1/2)^(n-1))
令Tn=1*(1/2)^0+4*(1/2)^1 + 7 * (1/2)^2 +...+(3n-2) (1/2)^(n-1)
1/2Tn= 1*(1/2)^1 +4* (1/2)^2 +7* (1/2)^3+...+(3n-5) (1/2)^(n-1)+(3n-2) (1/2)^(n)
两式相减 ,得 1/2*Tn=1*(1/2)^0+3*(1/2)^1+ 3*(1/2)^2+...+3*(1/2)^(n-1)- (3n-2) (1/2)^(n)
1/2*Tn=1+3*[(1/2)^1+(1/2)^2+...+(1/2)^(n-1)]-(3n-2) (1/2)^(n)
1/2*Tn=4 - (3n+4)*(1/2)^n
Tn= 8- (6n+8)*(1/2)^n
Sn= (2+8+14+,+6n-4) - Tn
=3n^2 -n - 8+(6n+8)*(1/2)^n ,n∈N*
An+2=3/2An+1-1/2An-----(An+2)-(An+1)=1/2((An+1)-(An))
所以(dn+1)=1/2dn;
d1=-1/3,所以dn=-1/3乘(1/2)的(n-1)次方;