∵an=an-1+1/n(n+1)
∴an-an-1=1/n-1/(n+1)
an-1-an-2=1/(n-1)-1/n
… … …
a2 -a1 =1-1/2
上述各式相加得:an-a1=1-1/(n+1)=n/(n+1)
∴an=1+n/(n+1)=(2n+1)/(n+1)
已知数列an中,a1=1,对任意自然数n都有an=an-1+1/n(n+1),求an的通项
已知数列an中,a1=1,对任意自然数n都有an=an-1+1/n(n+1),求an的通项
数学人气:580 ℃时间:2019-10-10 04:17:58
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