Sn=n(2n-1)an a1=1/3求数列的Sn,an
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Sn=n(2n-1)ana1=1/3求数列的Sn,anSn=n(2n-1)ana1=1/3求数列的Sn,anSn=n(2n-1)ana1=1/3求数列的Sn,an当n=1时,可知S1=a1=1/3.当
Sn=n(2n-1)an a1=1/3求数列的Sn,an
Sn=n(2n-1)an a1=1/3求数列的Sn,an
Sn=n(2n-1)an a1=1/3求数列的Sn,an
当n=1时,可知S1=a1=1/3.
当n>1时,因为Sn=n(2n-1)an
Sn-1=(n-1)[2(n-1)-1]an-1
由于 Sn-Sn-1=an
可知 n(2n-1)an-(n-1)[2(n-1)-1]an-1=an
化简得 an=[(2n-3)/(2n+1)]an-1
an-1=[(2n-5)/(2n-1)]an-2
an-2=[(2n-7)/(2n-3)]an-3
······
a3=(3/7)a2
a2=(1/5)a1
a1=1/3
然后把左边的例如an-1、an-2等等带到an中,
化简的 an=1/(2n+1)(2n-1)
所以Sn=[n(2n-1)]/(2n+1)(2n-1)=n/(2n+1)
把原式当(1);然后令n=n-1带入(1)式得(2);然后(1)-(2)得an/an-1=(2n-3)/(2n+1);然后迭乘得an
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