已知数列{an},a1=1,a2=2,且有2a(n+2)=a(n-1)+ann属于正整数（1）令bn=a(n+1)-an,求证数列{bn}是等差数列（2）求an的通项公式（3）求an的前n项和Sn

Question

已知数列{an},a1=1,a2=2,且有2a(n+2)=a(n-1)+an
n属于正整数（1）令bn=a(n+1)-an,求证数列{bn}是等差数列（2）求an的通项公式（3）求an的前n项和Sn

Answer 1

2a(n+2)=a(n+1)+a(n),
2a(n+2)-2a(n+1)=a(n)-a(n+1),
a(n+2)-a(n+1)=(-1/2)[a(n+1)-a(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.
a(n+1)-a(n)=(-1/2)^(n-1),
(-2)^na(n+1) - (-2)^na(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)=(-2)^(n-1)a(n) + 2/3,
(-2)^(n-1)a(n) = (5/3)(-2)^(n-1) - 2/3,
a(n) = 5/3 - (2/3)(-1/2)^(n-1)
s(n) = 5n/3 - (2/3)[1-(-1/2)^n]/[1-(-1/2)]
=5n/3 - (2/3)[1-(-1/2)^n]/(3/2)
=5n/3 - (4/9)[1-(-1/2)^n]