已知数列{an}的前n项和为Sn=n²+2n(1)求数列的通项公式an(2)Tn=1/a1a2+1/a2a3+
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已知数列{an}的前n项和为Sn=n²+2n(1)求数列的通项公式an(2)Tn=1/a1a2+1/a2a3+1/a3a4+...+1/ana
(1)Sn=n2+2n
an=Sn-S(n-1)
=n2+2n-[(n-1)2+2(n-1)]
=2n+1
(2)1/ana(n+1)
=1/(2n+1)(2n+3)
=[1/(2n+1)-1/(2n+3)]/2
∴Tn=[1/3-1/5+1/5-1/7+1/7-1/9+……+1/(2n+1)-1/(2n+3)]/2
=[1/3-1/(2n+3)]/2
=n/[3(2n+3)]
an=Sn-S(n-1)
=n2+2n-[(n-1)2+2(n-1)]
=2n+1
(2)1/ana(n+1)
=1/(2n+1)(2n+3)
=[1/(2n+1)-1/(2n+3)]/2
∴Tn=[1/3-1/5+1/5-1/7+1/7-1/9+……+1/(2n+1)-1/(2n+3)]/2
=[1/3-1/(2n+3)]/2
=n/[3(2n+3)]
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