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.要写明方
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)
……