已知数列{An}的各项均为正数,前n项和为Sn,且满足2Sn=An²+n-4 1.求证{An}为等差数列2.求{An}的通项公式
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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=An²+n-4 1.求证{An}为等差数列2.求{An}的通项公式](/uploads/image/z/1270229-5-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9BAn%EF%BD%9D%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%26%23178%3B%2Bn-4+1.%E6%B1%82%E8%AF%81%EF%BD%9BAn%EF%BD%9D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%972.%E6%B1%82%EF%BD%9BAn%EF%BD%9D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
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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=An²+n-4 1.求证{An}为等差数列2.求{An}的通项公式
已知数列{An}的各项均为正数,前n项和为Sn,且满足2Sn=An²+n-4 1.求证{An}为等差数列
2.求{An}的通项公式
已知数列{An}的各项均为正数,前n项和为Sn,且满足2Sn=An²+n-4 1.求证{An}为等差数列2.求{An}的通项公式
1.
n=1时,2a1=2S1=a1²+1-4
a1²-2a1-3=0
(a1+1)(a1-3)=0
a1=-1(数列各项均为正,舍去)或a1=3
n≥2时,
2an=2Sn-2S(n-1)=an²+n-4-a(n-1)²-(n-1)+4=an²-a(n-1)²+1
an²-2an+1=a(n-1)²
(an -1)²=a(n-1)²
an-1=-a(n-1)或an-1=a(n-1)
an -1=-a(n-1)时,an+a(n-1)=1 a2+a1=1 a2=1-a1=1-3=-2