{an},{bn}都是各项为正的数列,对任意的自然数n,都有an,(bn)^2,a(n+1)成等差数列,(bn)^2,a(n+1),[b(n+1)]^2成等比数列(1)求证:{bn}是等差数列(2)若a1=1,b1=√2,Sn=1/a1+1/a2+…+1/an,求Sn
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![{an},{bn}都是各项为正的数列,对任意的自然数n,都有an,(bn)^2,a(n+1)成等差数列,(bn)^2,a(n+1),[b(n+1)]^2成等比数列(1)求证:{bn}是等差数列(2)若a1=1,b1=√2,Sn=1/a1+1/a2+…+1/an,求Sn](/uploads/image/z/1758839-23-9.jpg?t=%7Ban%7D%2C%7Bbn%7D%E9%83%BD%E6%98%AF%E5%90%84%E9%A1%B9%E4%B8%BA%E6%AD%A3%E7%9A%84%E6%95%B0%E5%88%97%2C%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84%E8%87%AA%E7%84%B6%E6%95%B0n%2C%E9%83%BD%E6%9C%89an%2C%28bn%29%5E2%2Ca%28n%2B1%29%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%28bn%29%5E2%2Ca%28n%2B1%29%2C%5Bb%28n%2B1%29%5D%5E2%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%7Bbn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%882%EF%BC%89%E8%8B%A5a1%3D1%2Cb1%3D%E2%88%9A2%2CSn%3D1%2Fa1%2B1%2Fa2%2B%E2%80%A6%2B1%2Fan%2C%E6%B1%82Sn)
{an},{bn}都是各项为正的数列,对任意的自然数n,都有an,(bn)^2,a(n+1)成等差数列,(bn)^2,a(n+1),[b(n+1)]^2成等比数列(1)求证:{bn}是等差数列(2)若a1=1,b1=√2,Sn=1/a1+1/a2+…+1/an,求Sn
{an},{bn}都是各项为正的数列,对任意的自然数n,都有an,(bn)^2,a(n+1)成等差数列,(bn)^2,a(n+1),[b(n+1)]^2成等比数列
(1)求证:{bn}是等差数列
(2)若a1=1,b1=√2,Sn=1/a1+1/a2+…+1/an,求Sn
{an},{bn}都是各项为正的数列,对任意的自然数n,都有an,(bn)^2,a(n+1)成等差数列,(bn)^2,a(n+1),[b(n+1)]^2成等比数列(1)求证:{bn}是等差数列(2)若a1=1,b1=√2,Sn=1/a1+1/a2+…+1/an,求Sn
an,(bn)^2,a(n+1)成等差数列,所以2×(bn)^2=an+a(n+1)……式1
(bn)^2,a(n+1),[b(n+1)]^2成等比数列 ,所以 a(n+1)^2=
(bn)^2×[b(n+1)]^2,即a(n+1)=(bn)×b(n+1),an=(bn)×b(n-1),代入式1有
2×(bn)^2=b(n-1)×bn+bn×b(n+1),则bn=b(n-1)+b(n+1),所以:{bn}是等差数列 成立
由an=(bn)×b(n-1),
1/an=1/(bn)×b(n-1)=[1/b(n-1)-1/(bn)]/(bn-b(n-1))=[1/b(n-1)-1/(bn)]/d,d是公差.
则Sn=1/a1+1/a2+…+1/an=[1/b0-1/(bn)]/d
若a1=1,b1=√2,b0=√2/2,d=√2/2,bn=b0+n*d=√2/2(n+1)所以Sn=2n/(n+1)