yanwang1104
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因为成等比数列
2Sn=an*a(n+1)
2S1=a1*a2 a2=2
2S(n+1)=a(n+1)*a(n+2)
2a(n+1)+2Sn=a(n+1)*a(n+2)
所以 2a(n+1)+a(n+1)*a(n+2)=a(n+1)*a(n+2) 化简,得2=a(n+2)-an
所以an是 4 2 6 4.即奇数项和偶数项是公差为2的等差数列.
当n=2k-1,an=2k+2 (k是N*)
当n=2k,an=2k 所以a(2k+2)=a(2k-1)
所以1/S1+1/S2+1/S3+……+1/S2n=1/a1+2/(a2*a3)+.2/(a2n*a(2n+1))
再裂项,奇数项和 :-1/a1+1/a2-1/a3+1/a4-.-1/a(2n-1)-a(2n)
偶数项和:1/2a2-1/2a3+.+1/2a(2n)-1/2a(2n+1)
相加,化简,得1/2a1+3/2a2-3/2a(2n-1)-1/2a(2n+1)=1/8+3/4-3/2(2n+2)-1/2(2n+4)
=7/8-3/(4n+4)-1/(4n+8)
1年前
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