己知数列A(n+1)=1/3A(n)+2n+5/3 n为正整数求A(n)关于n的通项公式注:()里的字母为下脚标.要写明
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/08/19 11:36:30
己知数列A(n+1)=1/3A(n)+2n+5/3 n为正整数求A(n)关于n的通项公式注:()里的字母为下脚标.要写明方
![己知数列A(n+1)=1/3A(n)+2n+5/3 n为正整数求A(n)关于n的通项公式注:()里的字母为下脚标.要写明](/uploads/image/z/18182601-9-1.jpg?t=%E5%B7%B1%E7%9F%A5%E6%95%B0%E5%88%97A%28n%2B1%29%3D1%2F3A%28n%29%2B2n%2B5%2F3+n%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%E6%B1%82A%28n%29%E5%85%B3%E4%BA%8En%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E6%B3%A8%EF%BC%9A%28%29%E9%87%8C%E7%9A%84%E5%AD%97%E6%AF%8D%E4%B8%BA%E4%B8%8B%E8%84%9A%E6%A0%87.%E8%A6%81%E5%86%99%E6%98%8E)
方法1.
A(n+1)=(1/3)A(n)+2n+5/3
An=(1/3)A(n-1)+2(n-1)+5/3
A2=(1/3)A1+2+5/3=(1/3)A1+11/3,
A3=(1/3)A2+4+5/3=(1/3)A2+17/3=(1/9)A1+62/9
两式相减:
A(n+1)-An=(1/3)A(n)+2n+5/3-[(1/3)A(n-1)+2(n-1)+5/3]
=(1/3)[A(n)-A(n-1)]+2
设Bn=A(n+1)-An,B1=A2-A1=-(2/3)A1+11/3,B2=A3-A2=-(2/9)A1+29/9,
Bn=(1/3)B(n-1)+2
B(n-1)=(1/3)B(n-2)+2
两式相减:
Bn-B(n-1)=(1/3)B(n-1)+2-[(1/3)B(n-2)+2]
=(1/3)[B(n-1)-B(n-2)]
设Cn=B(n+1)-Bn,C1=B2-B1=(4/9)(A1-1)
C(n-1)=(1/3)C(n-2)]
Cn=C1*(1/3)^(n-1)
=(4/9)(A1-1)(1/3)^(n-1)
B(n+1)-Bn=Cn=(4/9)(A1-1)(1/3)^(n-1)
Bn-B(n-1)=(4/9)(A1-1)(1/3)^(n-2)
B(n-1)-B(n-2)=(4/9)(A1-1)(1/3)^(n-3)
……
B4-B3=(4/9)(A1-1)(1/3)^2
B3-B2=(4/9)(A1-1)(1/3)^1
B2-B1=(4/9)(A1-1)(1/3)^0
两边相加:
Bn-B1=(4/9)(A1-1)[(1/3)^(n-2)+(1/3)^(n-3)+(1/3)^(n-4)+……+(1/3)^3+(1/3)^2+(1/3)+1]
=(4/9)(A1-1)[1-(1/3)^(n-1)]/(1-1/3)
=(2/3)(A1-1)[1-(1/3)^(n-1)]
Bn=(2/3)(A1-1)[1-(1/3)^(n-1)]-(2/3)A1+11/3
A(n+1)-An=Bn=(2/3)(A1-1)[1-(1/3)^(n-1)]-(2/3)A1+11/3
An-A(n-1)=(2/3)(A1-1)[1-(1/3)^(n-2)]-(2/3)A1+11/3
按上述叠加法:
An-A1=(2/3)(A1-1){[1-(1/3)^(n-2)]+[1-(1/3)^(n-3)]+[1-(1/3)^(n-4)]+……+[1-(1/3)^2]+[1-(1/3)^1]+[1-(1/3)^0}+(n-1)[-(2/3)A1+11/3]
=(2/3)(A1-1){n-1-[(1/3)^(n-2)+(1/3)^(n-3)+(1/3)^(n-4)]+……+(1/3)^2+(1/3)^1+(1/3)^0]}-(n-1)[(2/3)A1-11/3]
=(2/3)(A1-1){n-1-[1-(1/3)^(n-1)]/(1-1/3)}-(n-1)[(2/3)A1-11/3]
=(2/3)(A1-1){n-1-[3/2-(3/2)(1/3)^(n-1)]}-(n-1)[(2/3)A1-11/3]
=(2/3)(A1-1)[n-5/2+(3/2)(1/3)^(n-1)]-(n-1)[(2/3)A1-11/3]
=(1/3)(A1-1)[2n-5+(1/3)^(n-2)]-(1/3)(n-1)(2A1-11)
An-A1=(1/3)(A1-1)[2n-5+(1/3)^(n-2)]-(1/3)(n-1)(2A1-11)
An=(1/3)(A1-1)[2n-5+(1/3)^(n-2)]-(1/3)(n-1)(2A1-11)+A1
=(2/3)(A1-1)n-(5/3)(A1-1)+(1/3)(A1-1)(1/3)^(n-2)-(1/3)(2A1-11)n+(1/3)(2A1-11)+A1
=3n-(5/3)(A1-1)+(A1-1)(1/3)^(n-1)+(5/3)A1-11/3
=3n+(A1-1)(1/3)^(n-1)-2
=(A1-1)(1/3)^(n-1)+3n-2 .
方法2.
A(n+1)=(1/3)A(n)+2n+5/3
3A(n+1)=A(n)+6n+5
3[A(n+1)-3n]=[A(n)-3n]+5
3[A(n+1)-3(n+1)]=[A(n)-3n]-4
设Bn=A(n)-3n,B1=A1-3
3B(n+1)=B(n)-4
3Bn=B(n-1)-4
3[B(n+1)-Bn]=B(n)-B(n-1)
B(n+1)-Bn=(B2-B1)(1/3)^(n-1)
……
A(n+1)=(1/3)A(n)+2n+5/3
An=(1/3)A(n-1)+2(n-1)+5/3
A2=(1/3)A1+2+5/3=(1/3)A1+11/3,
A3=(1/3)A2+4+5/3=(1/3)A2+17/3=(1/9)A1+62/9
两式相减:
A(n+1)-An=(1/3)A(n)+2n+5/3-[(1/3)A(n-1)+2(n-1)+5/3]
=(1/3)[A(n)-A(n-1)]+2
设Bn=A(n+1)-An,B1=A2-A1=-(2/3)A1+11/3,B2=A3-A2=-(2/9)A1+29/9,
Bn=(1/3)B(n-1)+2
B(n-1)=(1/3)B(n-2)+2
两式相减:
Bn-B(n-1)=(1/3)B(n-1)+2-[(1/3)B(n-2)+2]
=(1/3)[B(n-1)-B(n-2)]
设Cn=B(n+1)-Bn,C1=B2-B1=(4/9)(A1-1)
C(n-1)=(1/3)C(n-2)]
Cn=C1*(1/3)^(n-1)
=(4/9)(A1-1)(1/3)^(n-1)
B(n+1)-Bn=Cn=(4/9)(A1-1)(1/3)^(n-1)
Bn-B(n-1)=(4/9)(A1-1)(1/3)^(n-2)
B(n-1)-B(n-2)=(4/9)(A1-1)(1/3)^(n-3)
……
B4-B3=(4/9)(A1-1)(1/3)^2
B3-B2=(4/9)(A1-1)(1/3)^1
B2-B1=(4/9)(A1-1)(1/3)^0
两边相加:
Bn-B1=(4/9)(A1-1)[(1/3)^(n-2)+(1/3)^(n-3)+(1/3)^(n-4)+……+(1/3)^3+(1/3)^2+(1/3)+1]
=(4/9)(A1-1)[1-(1/3)^(n-1)]/(1-1/3)
=(2/3)(A1-1)[1-(1/3)^(n-1)]
Bn=(2/3)(A1-1)[1-(1/3)^(n-1)]-(2/3)A1+11/3
A(n+1)-An=Bn=(2/3)(A1-1)[1-(1/3)^(n-1)]-(2/3)A1+11/3
An-A(n-1)=(2/3)(A1-1)[1-(1/3)^(n-2)]-(2/3)A1+11/3
按上述叠加法:
An-A1=(2/3)(A1-1){[1-(1/3)^(n-2)]+[1-(1/3)^(n-3)]+[1-(1/3)^(n-4)]+……+[1-(1/3)^2]+[1-(1/3)^1]+[1-(1/3)^0}+(n-1)[-(2/3)A1+11/3]
=(2/3)(A1-1){n-1-[(1/3)^(n-2)+(1/3)^(n-3)+(1/3)^(n-4)]+……+(1/3)^2+(1/3)^1+(1/3)^0]}-(n-1)[(2/3)A1-11/3]
=(2/3)(A1-1){n-1-[1-(1/3)^(n-1)]/(1-1/3)}-(n-1)[(2/3)A1-11/3]
=(2/3)(A1-1){n-1-[3/2-(3/2)(1/3)^(n-1)]}-(n-1)[(2/3)A1-11/3]
=(2/3)(A1-1)[n-5/2+(3/2)(1/3)^(n-1)]-(n-1)[(2/3)A1-11/3]
=(1/3)(A1-1)[2n-5+(1/3)^(n-2)]-(1/3)(n-1)(2A1-11)
An-A1=(1/3)(A1-1)[2n-5+(1/3)^(n-2)]-(1/3)(n-1)(2A1-11)
An=(1/3)(A1-1)[2n-5+(1/3)^(n-2)]-(1/3)(n-1)(2A1-11)+A1
=(2/3)(A1-1)n-(5/3)(A1-1)+(1/3)(A1-1)(1/3)^(n-2)-(1/3)(2A1-11)n+(1/3)(2A1-11)+A1
=3n-(5/3)(A1-1)+(A1-1)(1/3)^(n-1)+(5/3)A1-11/3
=3n+(A1-1)(1/3)^(n-1)-2
=(A1-1)(1/3)^(n-1)+3n-2 .
方法2.
A(n+1)=(1/3)A(n)+2n+5/3
3A(n+1)=A(n)+6n+5
3[A(n+1)-3n]=[A(n)-3n]+5
3[A(n+1)-3(n+1)]=[A(n)-3n]-4
设Bn=A(n)-3n,B1=A1-3
3B(n+1)=B(n)-4
3Bn=B(n-1)-4
3[B(n+1)-Bn]=B(n)-B(n-1)
B(n+1)-Bn=(B2-B1)(1/3)^(n-1)
……
数列{a(n)}的前n项和为S(n),a(1)=1,a(n+1)=2S(n)(∈正整数N).求数列{a(n)}的通项公式
已知数列 {a(n)} 的通项公式为a(n)=1/(n²+2n),求数列 {a(n)}前n项和
已知数列{an}的通项公式a=2n,n为偶数,1-3n,n为奇数,求该数列的前100项和
已知数列{an}满足2a(n+1)(下脚标)=an+a(n+2)(下脚标)(n∈N+)它的前n项和为Sn,且a3=10,
设数列{a(n)}的前n项和为Sn,已知ba(n)-2^n=(b-1)Sn求{a(n)}的通项公式
求数列a(n+1)=ban+c^n,(b,c为常数,n为正整数)通项公式求法
在数列{a(n)}中,a1=3,a(n+1)=a(n)^2,n是正整数,求该数列的通项
若数列a(n)的递推关系满足a(n+1)/a(n)=(n+2)/n 求a(n)的通项公式
aˇn+1=2*aˇn+3,求数列{aˇn}的通项公式?
已知数列{An}的通项公式An=(1+2+3+……+n)/(n) ,Bn=1/(An*A(n+1))(n为正整数),求数
设数列An的前n项和为Sn,已知a(1)+2a(2)+3a(3)+…+na(n)=(n-1)Sn+2n(n为正整数).求
已知数列a[n]通项公式为a[n]=2^n/n,求前n项和