Question

a1=1,an=4an-1+3^n-2n+1,求通项公式

Answer 1

构造等比数列：

an-2n/3=4(an-1 - 2(n-1)/3)-5/3+3^n

令bn=an-2n/3

bn=4bn-1+3^n-5/3

令3^n -5/3 =Tn-1

bn=4bn-1+Tn-1

bn=4(4bn-2+Tn-2)+Tn-1

=4^2bn-2+4Tn-2+Tn-1

=4^2(4bn-3+Tn-3)+4Tn-2+Tn-1

=4^3bn-3+4^2Tn-3+4Tn-2+Tn-1

.

bn=4^(n-1)b1+4^(n-2)T1+.4^2Tn-3+Tn-2+Tn-1

b1=1/3 T1=3^2-5/3

bn=1/3*4^(n-1)+ {4^(n-2)3^2+.+4^2*3^(n-2)+4*3^(n-1)+3^n} - 5/3{4^(n-2)+.+1}

bn={ 3^n - 4^(n-2)3^2*(4/3) / 1- (4/3) } + 1/3*4(n-1) - 5/3 {4^(n-1) - 1 / 3}

an=bn+2n/3

an={ 4^(n-1)*9 - 3^(n+1) } + 1/3*4(n-1) - 5/3 {4^(n-1) -1 / 3}+ 2n/3

做完以后我觉得这题有点变态.用了好多种技巧,计算量还非常大.