已知数列an的前n项和是Sn=n^2,则1/a1a2+1/a2a3+...+1/a(n-1)an=?
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已知数列an的前n项和是Sn=n^2,则1/a1a2+1/a2a3+...+1/a(n-1)an=?
an=Sn-S(n-1)
=n^2-(n-1)^2
=2n-1
这是个等差奇数列
1/(an*a(n+1))
=1/(a(n+1)-an)*(1/an-1/a(n+1))
=1/2(1/an-1/a(n+1))
所以
1/a1a2+1/a2a3+...+1/a(n-1)an
=1/2(1/a1-1/a2+1/a2-1/a3+1/a3-1/a4+.+1/a(n-1)-1/an)
=1/2(1/1-1/(2n-1))
=(n-1)/(2n-1)
不懂的欢迎追问,
=n^2-(n-1)^2
=2n-1
这是个等差奇数列
1/(an*a(n+1))
=1/(a(n+1)-an)*(1/an-1/a(n+1))
=1/2(1/an-1/a(n+1))
所以
1/a1a2+1/a2a3+...+1/a(n-1)an
=1/2(1/a1-1/a2+1/a2-1/a3+1/a3-1/a4+.+1/a(n-1)-1/an)
=1/2(1/1-1/(2n-1))
=(n-1)/(2n-1)
不懂的欢迎追问,
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