已知数列an满足a1=1,a2=2,an+1=an+an-1/2令bn=an+1-an,证明bn是等比数列n属于N*求an通项公式

Question

已知数列an满足a1=1,a2=2,an+1=an+an-1/2
令bn=an+1-an,证明bn是等比数列n属于N*
求an通项公式

Answer 1

a(1)=1,a(2)=2,
a(n+2)=[a(n+1)+a(n)]/2,
a(n+2)-a(n+1)=[a(n)-a(n+1)]/2=(-1/2)[a(n+1)-a(n)],
b(n+1)=a(n+2)-a(n+1)=(-1/2)[a(n+1)-a(n)]=(-1/2)b(n),
{b(n)=a(n+1)-a(n)}是首项为a(2)-a(1)=1,公比为(-1/2)的等比数列.
b(n)=a(n+1)-a(n)=(-1/2)^(n-1),
a(n+1) = a(n) + (-1/2)^(n-1),
(-2)^na(n+1) = (-2)(-2)^(n-1)a(n) - 2,
c(n) = (-2)^(n-1)a(n),
c(n+1)=(-2)^na(n+1)=(-2)(-2)^(n-1)a(n) - 2 = -2c(n)-2,
c(n+1)+2/3 = -2c(n)-2+2/3 = -2c(n) - 4/3 = (-2)[c(n) + 2/3],
{c(n)+2/3}是首项为c(1)+2/3=a(1)+2/3=5/3,公比为（-2）的等比数列.
c(n)+2/3 = (5/3)(-2)^(n-1),
c(n) = (5/3)(-2)^(n-1) - 2/3 = (-2)^(n-1)a(n),
a(n) = 5/3 - (2/3)(-1/2)^(n-1)