Question

已知数列﹛an﹜满足a1=0且a(n+1)=2an+n (n∈N*)

Answer 1

（1）a2=2a1+1=1, a3=2a2+1=3

由a(n+1)=2an+n可得：

a(n+1)-an=an+n, n=a(n+1)-2an

a(n+2)-a(n+1)=a(n+1)+n+1=a(n+1)+a(n+1)-2an+1=2[a(n+1)-an]+1

（2）bn=a(n+1)-an, b1=a2-a1=1

b(n+1)=a(n+2)-a(n+1) =2[a(n+1)-an]+1=2bn+1

设Cn=bn+1, bn=Cn-1, c1=b1+1=2

C(n+1)-1=2(Cn-1)+1=2Cn-1

C(n+1)=2Cn

Cn为首相=2,等比=2的等比数列

Cn=2^n

Bn=Cn-1=2^n-1

（2）bn=a(n+1)-an=2^n-1

a(n+1)=2^n+an

an=2^(n-1)+a(n-1)-1

a2=2^1-1

a3=2^2+(2^1-1)-1=2^2+2^1-2

a4=2^3+(2^2+2^1-2)-1=2^3+2^2+2^1-3

.

an=2^(n-1)+a(n-1)-1=2^(n-1)+2^(n-2)+.+2^1-(n-1)

=2[2^(n-1)-1]/(2-1) - (n-1)

=2[2^(n-1)-1] - (n-1)

=2^n-2-n+1

=2^n-n-1