已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2*a3=45,a1+a4=14(1)求数列{an}通项公式.我会,答案是an=4n-3(2)设数列bn=1/an*a(n+1)qi求数列{bn}前n项和Sn第(2)题答案是sn=n/4n+1
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![已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2*a3=45,a1+a4=14(1)求数列{an}通项公式.我会,答案是an=4n-3(2)设数列bn=1/an*a(n+1)qi求数列{bn}前n项和Sn第(2)题答案是sn=n/4n+1](/uploads/image/z/1087098-42-8.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2C%E5%85%AC%E5%B7%AEd%3E0%2C%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%3Aa2%2Aa3%3D45%2Ca1%2Ba4%3D14%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.%E6%88%91%E4%BC%9A%2C%E7%AD%94%E6%A1%88%E6%98%AFan%3D4n-3%EF%BC%882%EF%BC%89%E8%AE%BE%E6%95%B0%E5%88%97bn%3D1%2Fan%2Aa%28n%2B1%29qi%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E7%AC%AC%EF%BC%882%EF%BC%89%E9%A2%98%E7%AD%94%E6%A1%88%E6%98%AFsn%3Dn%2F4n%2B1)
已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2*a3=45,a1+a4=14(1)求数列{an}通项公式.我会,答案是an=4n-3(2)设数列bn=1/an*a(n+1)qi求数列{bn}前n项和Sn第(2)题答案是sn=n/4n+1
已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2*a3=45,a1+a4=14
(1)求数列{an}通项公式.我会,答案是an=4n-3
(2)设数列bn=1/an*a(n+1)qi求数列{bn}前n项和Sn
第(2)题答案是sn=n/4n+1
已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2*a3=45,a1+a4=14(1)求数列{an}通项公式.我会,答案是an=4n-3(2)设数列bn=1/an*a(n+1)qi求数列{bn}前n项和Sn第(2)题答案是sn=n/4n+1
1、
略
2、
a(n+1)=4n+1
bn=1/4*4/(4n-3)(4n+1)
=1/4*[(4n+1)-(4n-3)]/[(4n-3)(4n+1)]
=1/4*{(4n+1)/[(4n-3)(4n+1)]-(4n-3)/[(4n-3)(4n+1)]}
=1/4*[1/(4n-3)-1/(4n+1)]
所以Sn=1/4*[1/1-1/5+1/5-1/9+1/9-1/13+……+1/(4n-3)-1/(4n+1)]
==1/4*[1-1/(4n+1)]
=n/(4n+1)
bn=1/(4n-7)*(4n-3)=1/4[1/(4n-7)-1/(4n-3)]
bn=b1+b2+^^^^^bn
=1/4(-1/3-1+1-1/5+1/5-1/9+1/9^^^^-1/(4n-3)前后相消;就得出结果了。
我不太喜欢算数,你自己算咯。这是一种题型,记着……