设a1=1,a2=5/3,a(n+2)=(5/3)[a(n+1)-(2/3)an(n=1,2,3…).
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设a1=1,a2=5/3,a(n+2)=(5/3)[a(n+1)-(2/3)an(n=1,2,3…).
(1)令bn=a(n+1)-an(n=1,2,…),求数列{bn}的通项公式;
(2)求数列{nan}的前n项和Sn
(1)令bn=a(n+1)-an(n=1,2,…),求数列{bn}的通项公式;
(2)求数列{nan}的前n项和Sn
![设a1=1,a2=5/3,a(n+2)=(5/3)[a(n+1)-(2/3)an(n=1,2,3…).](/uploads/image/z/821837-29-7.jpg?t=%E8%AE%BEa1%3D1%2Ca2%3D5%2F3%2Ca%28n%2B2%29%3D%285%2F3%29%5Ba%28n%2B1%29-%282%2F3%29an%28n%3D1%2C2%2C3%E2%80%A6%EF%BC%89.)
1.a(n+2)-a(n+1)=2/3 [a(n+1)-an]
[a(n+2)-a(n+1)]/[a(n+1)-an]=2/3
a(n+1)-an=(2/3)^n
bn=(2/3)^n
2.
a(n+1)-an=(2/3)^n
a2-a1=2/3
a3-a2=(2/3)^2
a4-a3=(2/3)^3
……
……
an-a(n-1)=(2/3)^(n-1)
叠加得
an-a1=2-(2/3)^(n-1)
an=3-(2/3)^(n-1)
nan=3n-n(2/3)^(n-1)
令Cn=n(2/3)^(n-1) 和为Tn
Tn=1*(2/3)^0+2*(2/3)^1+3*(2/3)^2+……+n*(2/3)^(n-1)
2/3Tn=1*(2/3)^1+2*(2/3)^2+3*(2/3)^3+……+(n-1)(2/3)^(n-1)+n*(2/3)^n
相减
1/3Tn=1*(2/3)^0+1*(2/3)^1+1*(2/3)^2+……+1*(2/3)^(n-1)-n*(2/3)^n
=3*[1-(2/3)^n]-n*(2/3)^n
Tn=9*[1-(2/3)^n]-3n*(2/3)^n
Sn=(1+3n)*n/2-Tn=(1+3n)*n/2-3*[1-(2/3)^n]+n*(2/3)^n
[a(n+2)-a(n+1)]/[a(n+1)-an]=2/3
a(n+1)-an=(2/3)^n
bn=(2/3)^n
2.
a(n+1)-an=(2/3)^n
a2-a1=2/3
a3-a2=(2/3)^2
a4-a3=(2/3)^3
……
……
an-a(n-1)=(2/3)^(n-1)
叠加得
an-a1=2-(2/3)^(n-1)
an=3-(2/3)^(n-1)
nan=3n-n(2/3)^(n-1)
令Cn=n(2/3)^(n-1) 和为Tn
Tn=1*(2/3)^0+2*(2/3)^1+3*(2/3)^2+……+n*(2/3)^(n-1)
2/3Tn=1*(2/3)^1+2*(2/3)^2+3*(2/3)^3+……+(n-1)(2/3)^(n-1)+n*(2/3)^n
相减
1/3Tn=1*(2/3)^0+1*(2/3)^1+1*(2/3)^2+……+1*(2/3)^(n-1)-n*(2/3)^n
=3*[1-(2/3)^n]-n*(2/3)^n
Tn=9*[1-(2/3)^n]-3n*(2/3)^n
Sn=(1+3n)*n/2-Tn=(1+3n)*n/2-3*[1-(2/3)^n]+n*(2/3)^n
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