已知等差数列〔an〕满足a1=1,an+1

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已知等差数列〔an〕满足a1=1,an+1
已知等差数列〔an〕满足a1=1,an+1

已知等差数列〔an〕满足a1=1,an+1
令公差为d,则a2=d+1,a3=1+2d,所以
(a1+1)(a3+3)=(a2+1)^2
4d+8=d^2+4d+4
d^2=4
又an+11(an*an+1)=(1/an-1/an+1)/d
所以Sn=[(1/a1-1/a2)+(1/a2-1/a3).+(1/an-1/an+1)]/d
=(1/a1-1/an+1)/d
=[1-1/(1-2n)]/d
=n/(1-2n)

令公差为d,则a2=d+1,a3=1+2d，所以
(a1+1)(a3+3)=(a2+1)^2
4d+8=d^2+4d+4
d^2=4
d=正负2
又an+11(an*an+1)=(1/an-1/an+1)/d
所以Sn=[(1/a1-1/a2)+(1/a2-1/a3).......+(1/an-1/an+1)]/d
=(1/a1-1/an+1)/d
=[1-1/(1-2n)]/d
=n/(1-2n)

(2+d)^2=2(4+2d) d<0 d=-2
an=1-2(n-1)=3-2n
1/an =1/(3-2n)
1/an＊an+1=1/2(1/an+1-1/an)

此题早几年我会，现在早都还给老师了。

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