数列{an}中a1=2,an+1=2^nan (1)求证:a1/2,a2,a3成等比数列(2)求{an}的通项公式
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数列{an}中a1=2,an+1=2^nan (1)求证:a1/2,a2,a3成等比数列(2)求{an}的通项公式
![数列{an}中a1=2,an+1=2^nan (1)求证:a1/2,a2,a3成等比数列(2)求{an}的通项公式](/uploads/image/z/17978127-15-7.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%ADa1%3D2%2Can%2B1%3D2%5Enan+%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9Aa1%2F2%2Ca2%2Ca3%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%882%EF%BC%89%E6%B1%82%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
a(n+1) =2^n.an
a1/2 = 1
a2 = 2^1.a1 = 4
a3 = 2^2 .a2 = 16
=>a1/2,a2,a3成等比数列
a(n+1) =2^n.an
loga(n+1) = logan + n
loga(n+1) - logan = n
logan - loga1 = 1+2+3+...+(n-1)
logan =n(n-1)/2 +1
=(n^2-n+2)/2
an = 2^[(n^2-n+2)/2]
再问: 第二问怎么这么复杂,还有log.....
再答: 这一类,一定要用log来作!
a1/2 = 1
a2 = 2^1.a1 = 4
a3 = 2^2 .a2 = 16
=>a1/2,a2,a3成等比数列
a(n+1) =2^n.an
loga(n+1) = logan + n
loga(n+1) - logan = n
logan - loga1 = 1+2+3+...+(n-1)
logan =n(n-1)/2 +1
=(n^2-n+2)/2
an = 2^[(n^2-n+2)/2]
再问: 第二问怎么这么复杂,还有log.....
再答: 这一类,一定要用log来作!
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