an=1,a(n+1)=2(an²+an).求an的通项公式.
1个回答

因为 a(n+1)=(an +2)/an,

所以 a(n+1) +1 =2(an +1)/an

1/[a(n+1) +1] =an/[2(an +1)]

1/[a(n+1) +1] =(an +1 -1)/[2(an +1)]=1/2 -1/[2(an +1)]

令 bn=1/(an +1),则

b(n+1) =-(1/2)bn +1/2

b(n+1) -1/3 =-(1/2)(bn -1/3)

而b1 - 1/3=1/(a1+1) -1/3=1/6,

所以 {bn -1/3}是首项为1/6,公比为-1/2的等比数列,

所以 bn -1/3 =(1/6)(-1/2)^(n-1)=(-1/3)(-1/2)^n,

bn=(1/3)[1 -(-1/2)^n]

an=1/bn -1 =3/[1 -(-1/2)^n ] -1

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