Question

数列{An}中,A1=1,S(n+1)=4An+2,用数学归纳法证明：An=(3n-1)*2^(n-2).

Answer 1

n=1,a1=1,成立

n=2,a2=s2-a1=5,成立

设n=k-1,n=k,成立

ak=(3k-1)*2^(k-2)

a(k-1)=(3k-4)*2^(k-3)

则a(k+1)=S(n+1)-S(n)

=4an+2-(4a(n-1)+2)

=4(an-a(n-1))

＝4*[(3k-1)*2^(k-2)-(3k-4)*2^(k-3)]

=4*(3k+2)*2^(k-3)

=[3(k+1)-1]2[(k+1)-2]

也成立

原命题得证