已知数列{an}的前n项和为Sn,且Sn=2–(2/n+1)an.
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(1)Sn=2-(2/n+1)an,①
n=1时a1=2-3a1,a1=1/2.
n>2时S<n-1>=2-[2/(n-1)+1]a<n-1>,②
①-②,an=-(2/n+1)an+[2/(n-1)+1]a<n-1>,
an/n=(1/2)*a<n-1>/(n-1)=……=(1/2)^(n-1)*a1=2^(-n),
∴an=n*2^(-n).
(2)bn=n/2,
1/[bn*b<n+2>]=4/[n(n+2)]=2[1/n-1/(n+2)],
∴原式=2[1-1/3+1/2-1/4+1/3-1/5+……+1/n-1/(n+2)]
=2[1+1/2-1/(n+1)-1/(n+2)]
=3-2(2n+3)/[(n+1)(n+2)].
n=1时a1=2-3a1,a1=1/2.
n>2时S<n-1>=2-[2/(n-1)+1]a<n-1>,②
①-②,an=-(2/n+1)an+[2/(n-1)+1]a<n-1>,
an/n=(1/2)*a<n-1>/(n-1)=……=(1/2)^(n-1)*a1=2^(-n),
∴an=n*2^(-n).
(2)bn=n/2,
1/[bn*b<n+2>]=4/[n(n+2)]=2[1/n-1/(n+2)],
∴原式=2[1-1/3+1/2-1/4+1/3-1/5+……+1/n-1/(n+2)]
=2[1+1/2-1/(n+1)-1/(n+2)]
=3-2(2n+3)/[(n+1)(n+2)].
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