各项均匀为正数的数列﹛an﹜的前n项和为Sn,满足4Sn=a²(n+1)-4n-1,n属于N*,a2,a5,a14构成等比1.证明a2=根号4a1+5 2.求数列﹛an﹜ 的通项公式
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![各项均匀为正数的数列﹛an﹜的前n项和为Sn,满足4Sn=a²(n+1)-4n-1,n属于N*,a2,a5,a14构成等比1.证明a2=根号4a1+5 2.求数列﹛an﹜ 的通项公式](/uploads/image/z/1160701-61-1.jpg?t=%E5%90%84%E9%A1%B9%E5%9D%87%E5%8C%80%E4%B8%BA%E6%AD%A3%E6%95%B0%E7%9A%84%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%BB%A1%E8%B6%B34Sn%3Da%26%23178%3B%EF%BC%88n%2B1%EF%BC%89-4n-1%2Cn%E5%B1%9E%E4%BA%8EN%2A%2Ca2%2Ca5%2Ca14%E6%9E%84%E6%88%90%E7%AD%89%E6%AF%941.%E8%AF%81%E6%98%8Ea2%3D%E6%A0%B9%E5%8F%B74a1%2B5+2.%E6%B1%82%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C+%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
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各项均匀为正数的数列﹛an﹜的前n项和为Sn,满足4Sn=a²(n+1)-4n-1,n属于N*,a2,a5,a14构成等比1.证明a2=根号4a1+5 2.求数列﹛an﹜ 的通项公式
各项均匀为正数的数列﹛an﹜的前n项和为Sn,满足4Sn=a²(n+1)-4n-1,n属于N*,a2,a5,a14构成等比
1.证明a2=根号4a1+5 2.求数列﹛an﹜ 的通项公式
各项均匀为正数的数列﹛an﹜的前n项和为Sn,满足4Sn=a²(n+1)-4n-1,n属于N*,a2,a5,a14构成等比1.证明a2=根号4a1+5 2.求数列﹛an﹜ 的通项公式
1、
4S1=a²2-4*1-1
S1=a1
a²2=4a1+5
a2=√(4a1+5)
2、
an=Sn-S(n-1)
4an=4Sn-4S(n-1)
=a²(n+1)-4n-1-[a²n-4(n-1)-1]
=a²(n+1)-4-a²n
a²n+4an+4=a²(n+1)
(an+2)²=a²(n+1)
各项均匀为正数
an+2=a(n+1)
a1+2=a2=√(4a1+5)
a1²+4a1+4=4a1+5
a1=1
{an}是一个首项是1,公差是2的等差数列
an=2n-1
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