首先,我觉得此题条件应该是:an=2a(n-1)+2^n+3
依据此:a2=2a1+2^2+3=1
a3=2a2+2^3+3=13
(2)证明:因bn=(an+3)/2^n,所以
bn-b(n-1)=(an+3)/2^n-(a(n-1)+3)/2^(n-1)
=(an+3)/2^n-2(a(n-1)+3)/2^n
=(an+3-2a(n-1)-6)/2^n
=(an-2a(n-1)-3)/2^n
=2^n/2^n
=1
所以,数列bn为公差为1的等差数列.
首先,我觉得此题条件应该是:an=2a(n-1)+2^n+3
依据此:a2=2a1+2^2+3=1
a3=2a2+2^3+3=13
(2)证明:因bn=(an+3)/2^n,所以
bn-b(n-1)=(an+3)/2^n-(a(n-1)+3)/2^(n-1)
=(an+3)/2^n-2(a(n-1)+3)/2^n
=(an+3-2a(n-1)-6)/2^n
=(an-2a(n-1)-3)/2^n
=2^n/2^n
=1
所以,数列bn为公差为1的等差数列.