0已知满足a1=1,a2=2,an+2=(an+1+an)/2 (1)令bn=an+1-an,证明是等比数列 (2)求an的通项公式.

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0已知满足a1=1,a2=2,an+2=(an+1+an)/2 (1)令bn=an+1-an,证明是等比数列 (2)求an的通项公式.
0已知满足a1=1,a2=2,an+2=(an+1+an)/2 (1)令bn=an+1-an,证明是等比数列 (2)求an的通项公式.

0已知满足a1=1,a2=2,an+2=(an+1+an)/2 (1)令bn=an+1-an,证明是等比数列 (2)求an的通项公式.
你把an+2=(an+1+an)/2变形一下,就可以得出bn+1/bn =(an+2 - an+1)/(an+1 - an)= -1/2,然后看b1 = a2 - a1 =1不为0,所以bn不就是首项为1,公比为-1/2的等比数列么?
bn = (-1/2)^(n-1) (表示(-1/2)的(n-1)次方),然后
a2 - a1 = (-1/2)^(1-1)
a3 - a2 =(-1/2)^(2-1)

an - an-1 = (-1/2)^(n-1-1)
然后相加,不就是an - a1 = (2/3)[1 - (-1/2)^(n-1)],an不就算出来了么?

1,∵a(n+2)=[a(n+1)+an]/2
∴2a(n+2)=a(n+1)+an
∴2a(n+2)-2a(n+1)=-a(n+1)+an
即2[a(n+2)-a(n+1)]=-[a(n+1)-an]
也即2b(n+1)=-bn
∴b(n+1)/bn=-1/2,为常数
而b1=a2-...

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1,∵a(n+2)=[a(n+1)+an]/2
∴2a(n+2)=a(n+1)+an
∴2a(n+2)-2a(n+1)=-a(n+1)+an
即2[a(n+2)-a(n+1)]=-[a(n+1)-an]
也即2b(n+1)=-bn
∴b(n+1)/bn=-1/2,为常数
而b1=a2-a1=2-1=1
∴bn是以1为首项、-1/2为公比的等比数列
2,bn=(-1/2)^(n-1)=a(n+1)-an
∴an-a(n-1)=(-1/2)^(n-2)
a(n-1)-a(n-2)=(-1/2)^(n-3)
………………………………
a2-a1=(-1/2)^0
累加,得:an-a1=(-1/2)^(n-2)+(-1/2)^(n-3)+……+1
=[1-(-1/2)^(n-1)]/(1+1/2)
=2/3-2/3*(-1/2)^(n-1)
所以an=5/3-2/3*(-1/2)^(n-1)

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