一道数学题:已知数列an的前n项和为sn,满足an+2sns(n-1)=0(n≥2,n为正整数),a1=1/2
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一道数学题:已知数列an的前n项和为sn,满足an+2sns(n-1)=0(n≥2,n为正整数),a1=1/2
1,求sn,an,的表达式 2.若bn=2(1-n)an(n≥2,n为正整数),求证,b2^2+b3^2+.+bn^2<2/3
1,求sn,an,的表达式 2.若bn=2(1-n)an(n≥2,n为正整数),求证,b2^2+b3^2+.+bn^2<2/3
![一道数学题:已知数列an的前n项和为sn,满足an+2sns(n-1)=0(n≥2,n为正整数),a1=1/2](/uploads/image/z/4677873-33-3.jpg?t=%E4%B8%80%E9%81%93%E6%95%B0%E5%AD%A6%E9%A2%98%EF%BC%9A%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E6%BB%A1%E8%B6%B3an%2B2sns%28n-1%29%3D0%28n%E2%89%A52%2Cn%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%29%2Ca1%3D1%2F2)
(1)an+2sns(n-1)=0,an=sn-s(n-1)
化为1/sn-1/s(n-1)=2
数列1/sn为等差数列.公差d=2,s1=a1=1/2,有1/s1=2,所以1/sn=(n-1)*d+1/s1=2n
sn=1/(2n)
an=sn-s(n-1)=1/[2n(1-n)]
(2)bn=2(1-n)an=2(1-n)/[2n(1-n)]=1/n,
b2^2+b3^2+.+bn^2
=(1/2)^2+(1/3)^2+.+(1/n)^2
化为1/sn-1/s(n-1)=2
数列1/sn为等差数列.公差d=2,s1=a1=1/2,有1/s1=2,所以1/sn=(n-1)*d+1/s1=2n
sn=1/(2n)
an=sn-s(n-1)=1/[2n(1-n)]
(2)bn=2(1-n)an=2(1-n)/[2n(1-n)]=1/n,
b2^2+b3^2+.+bn^2
=(1/2)^2+(1/3)^2+.+(1/n)^2
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