已知数列{an}的各项均为正数,前n项和为Sn,且满足2Sn=an^2+n-4求{an}的通项公式及
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![已知数列{an}的各项均为正数,前n项和为Sn,且满足2Sn=an^2+n-4求{an}的通项公式及](/uploads/image/z/3023368-16-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94%E6%BB%A1%E8%B6%B32Sn%3Dan%5E2%2Bn-4%E6%B1%82%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E5%8F%8A)
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已知数列{an}的各项均为正数,前n项和为Sn,且满足2Sn=an^2+n-4求{an}的通项公式及
已知数列{an}的各项均为正数,前n项和为Sn,且满足2Sn=an^2+n-4求{an}的通项公式及
已知数列{an}的各项均为正数,前n项和为Sn,且满足2Sn=an^2+n-4求{an}的通项公式及
因为2Sn=an^2+n-4,所以2S(n-1)=a(n-1)²+n-1-4.
两式相减2an=an^2-a(n-1)²+1,a(n-1)²=an^2-2an+1=(an-1)²
因为各项都是正数,所以a (n-1)=a n - 1.令n=1,2a1=a1²+1-4,a1=3.
所以{an}是以a1=3为首项,d=1为公差的等差数列.
an=n+2.
设:an=a*n+b
2sn=(a1+an)*n
=(a*n+b+a+b)*n
=a*n^2+(a+2b)n
所以a*n^2+(a+2b)n=an^2+n-4
an^2=a*n^2+(a+2b-1)n+4
有a1=a+b得a=1,b=2
所以an=n+2