已知数列a1=1/2,Sn=(n+1)(2n+1)an,则an=
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已知数列a1=1/2,Sn=(n+1)(2n+1)an,则an=
Sn=(n+1)(2n+1)an
S(n-1)=n(2n-1)a(n-1)
相减:
an=Sn-S(n-1)=(n+1)(2n+1)an-n(2n-1)a(n-1)
[(n+1)(2n+1)-1]an=n(2n-1)a(n-1)
n(2n+3)an=n(2n-1)a(n-1)
an/a(n-1)=(2n-1)/(2n+3)
a(n-1)/a(n-2)=(2n-3)/(2n+1)
a(n-2)/a(n-3)=(2n-5)/(2n-1)
……
a4/a3=7/11
a3/a2=5/9
a2/a1=3/7
叠乘:
an/a1=3*5/[(2n+3)(2n+1)]
an=(15/2)/[(2n+3)(2n+1)]
S(n-1)=n(2n-1)a(n-1)
相减:
an=Sn-S(n-1)=(n+1)(2n+1)an-n(2n-1)a(n-1)
[(n+1)(2n+1)-1]an=n(2n-1)a(n-1)
n(2n+3)an=n(2n-1)a(n-1)
an/a(n-1)=(2n-1)/(2n+3)
a(n-1)/a(n-2)=(2n-3)/(2n+1)
a(n-2)/a(n-3)=(2n-5)/(2n-1)
……
a4/a3=7/11
a3/a2=5/9
a2/a1=3/7
叠乘:
an/a1=3*5/[(2n+3)(2n+1)]
an=(15/2)/[(2n+3)(2n+1)]
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