已知数列{an}满足:a1=1/2,an+2Sn*S(n-1)=0(n≥2).(1)求证{1/Sn}是等差数列;(2)求an的表达式;(3)若bn=2(1-n),求证(b2)^2+(b3)^2+…+(bn)^2<1.
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![已知数列{an}满足:a1=1/2,an+2Sn*S(n-1)=0(n≥2).(1)求证{1/Sn}是等差数列;(2)求an的表达式;(3)若bn=2(1-n),求证(b2)^2+(b3)^2+…+(bn)^2<1.](/uploads/image/z/1769353-25-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3%EF%BC%9Aa1%3D1%2F2%2Can%2B2Sn%2AS%28n-1%29%3D0%EF%BC%88n%E2%89%A52%EF%BC%89.%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%7B1%2FSn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82an%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F%EF%BC%9B%EF%BC%883%EF%BC%89%E8%8B%A5bn%3D2%281-n%29%2C%E6%B1%82%E8%AF%81%28b2%29%5E2%2B%28b3%29%5E2%2B%E2%80%A6%2B%28bn%29%5E2%EF%BC%9C1.)
已知数列{an}满足:a1=1/2,an+2Sn*S(n-1)=0(n≥2).(1)求证{1/Sn}是等差数列;(2)求an的表达式;(3)若bn=2(1-n),求证(b2)^2+(b3)^2+…+(bn)^2<1.
已知数列{an}满足:a1=1/2,an+2Sn*S(n-1)=0(n≥2).
(1)求证{1/Sn}是等差数列;
(2)求an的表达式;
(3)若bn=2(1-n),求证(b2)^2+(b3)^2+…+(bn)^2<1.
已知数列{an}满足:a1=1/2,an+2Sn*S(n-1)=0(n≥2).(1)求证{1/Sn}是等差数列;(2)求an的表达式;(3)若bn=2(1-n),求证(b2)^2+(b3)^2+…+(bn)^2<1.
an=Sn-S(n-1)
an+2Sn*S(n-1)=0
带入后两边都除以Sn*S(n-1)
得 1/Sn-1/S(n-1)=2
所以{1/Sn}是等差数列 公差=2 首项1/S1=2
1/Sn=2n
Sn=1/2n
an=Sn-S(n-1) n≥2
=1/2n-1/2(n-1)
a1=1/2
(3)
这一问我就看不懂了
明明(b2)^2=4 然后(b3)^2 .(bn)^2 都大于0 它们相加怎么会小于1
请指正
1
an=sn-sn-1,所以sn-1-sn=2sn*sn-1
两边除以2sn-1sn,得1/sn-1/sn-1=2 所以1/sn是等差数列
2
1/sn=2n,所以sn=1/2n 即an=sn-sn-1=1/2n-1/2(n-1) n≥2 a1=1/2
3
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