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已知数列an的前n项和为Sn=n^2+2n,求和:1/(a1*a2)+1/(a2*a3)+...+1/(an*a(n+1))
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已知数列an的前n项和为Sn=n^2+2n,求和:1/(a1*a2)+1/(a2*a3)+...+1/(an*a(n+1))
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答案和解析
a1=S1=3,
n>1时,
an=Sn-S(n-1)=n^2+2n-(n-1)^2-2(n-1)=2n+1,
a1=2*1+1=3,所以通项公式为
an=2n-1
1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]
=1/(3*5)+1/(5*7)+1/(7*9)+.+1/[(2n+1)(2n+3)]
=1/3-1/5+1/5-1/7+1/7-1/9+.+1/(2n+1)-1/(2n+3)
=1/3-1/(2n+3)
=2n/[3(2n+3)]
1/(a1*a2)+1/(a2*a3)+...+1/(an*a(n+1))
=2n/[3(2n+3)]
n>1时,
an=Sn-S(n-1)=n^2+2n-(n-1)^2-2(n-1)=2n+1,
a1=2*1+1=3,所以通项公式为
an=2n-1
1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]
=1/(3*5)+1/(5*7)+1/(7*9)+.+1/[(2n+1)(2n+3)]
=1/3-1/5+1/5-1/7+1/7-1/9+.+1/(2n+1)-1/(2n+3)
=1/3-1/(2n+3)
=2n/[3(2n+3)]
1/(a1*a2)+1/(a2*a3)+...+1/(an*a(n+1))
=2n/[3(2n+3)]
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