已知数列{an}满足,a1=1,a2=2,an+2=(an十an+1)/2,n∈N.〈1〉令bn=an+1-an,证明:{bn}是等比数列:求{an...已知数列{an}满足,a1=1,a2=2,an+2=(an十an+1)/2,n∈N.〈1〉令bn=an+1-an,证明:{bn}是等比数列:求{an}的通项公式

2a[n+2]=an+a[n+1]同时-2a[n+1]
2(a[n+2]-a[n+1])=an-a[n+1]=-(a[n+1]-an)
2b[n+1]=-bn
b[n+1]/bn=-1/2 b1=1
bn=(-1/2)^(n-1)
an-a[n-1]=(-1/2)^(n-2)
..a2-a1=1
n-1个式子相加
an=[5+3(1/2)^(n-1)]/2

由a(n+2)=[an+a(n+1)]/2
a(n+2)-a(n+1)=[an-a(n+1)]/2
化b(n+1)=-1/2*bn(因bn=a(n+1)-an)
{bn}是等比数列，b1=1,公比-1/2，
bn=(-1/2)^(n-1)
an-a(n-1)=(-1/2)^(n-2)
……
a2-a1=1
n-1个式子累加an=[5+4(-1/2)^n]/3

a(n+2)=[an十a(n+1)]/2=a(n+2)-a(n+1)=[an-a(n+1)]/2
化b(n+1)=-1/2*bn(因bn=a(n+1)-an)
{bn}等比数列,b1=1,公比-1/2
则bn=(-1/2)^(n-1)
得an-a(n-1)=(-1/2)^(n-2)
……
n-1式子累加an=5/3-2[(-1/2)^(n-1)]/3