Question

数列中,sn=4an-1+1,且a1=1,若bn=an+1-2an,求证bn是等比

Answer 1

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

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

b(n)=a(n+1)-2a(n)

b(n)=2b(n-1)

s2=4a1+1=5

a2=s2-a1=4

b1=a2-2a1=2不为0

bn为等比数列,首项为2,公比为2,b(n)=2^n.