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已知数列{an}的前n项和为Sn,a1=1,an≠0,anan+1=λSn-1,其中λ为常数. (Ⅰ)证明:an+2-an=λ (Ⅱ)是否存在λ,使得{an}为等差数列?并说明理由.
答案:
(1)证明:由题设,anan+1=λSn-1,an+1an+2=λSn+1-1,
两式相减得an+1(an+2-an)=λan+1.
因为an+1≠0,所以an+2-an=λ.
(2)由题设,a1=1,a1a2=λS1-1,可得 a2=λ-1,
由(1)知,a3=λ+1.
若{an}为等差数列,则2a2=a1+a3,解得λ=4,故an+2-an=4.
由此可得{a2n-1}是首项为1,公差为4的等差数列,
a2n-1=4n-3;
{a2n}是首项为3,公差为4的等差数列,a2n=4n-1.
所以an=2n-1,an+1-an=2.
因此存在λ=4,使得数列{an}为等差数列.
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