求通项:a1=3,且a(n+1)=a²n(n∈正整数)答案说用迭代法求~迭代法是怎么算的啊?
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![求通项:a1=3,且a(n+1)=a²n(n∈正整数)答案说用迭代法求~迭代法是怎么算的啊?](/uploads/image/z/4032470-38-0.jpg?t=%E6%B1%82%E9%80%9A%E9%A1%B9%EF%BC%9Aa1%3D3%2C%E4%B8%94a%EF%BC%88n%2B1%EF%BC%89%3Da%26%23178%3Bn%EF%BC%88n%E2%88%88%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89%E7%AD%94%E6%A1%88%E8%AF%B4%E7%94%A8%E8%BF%AD%E4%BB%A3%E6%B3%95%E6%B1%82%7E%E8%BF%AD%E4%BB%A3%E6%B3%95%E6%98%AF%E6%80%8E%E4%B9%88%E7%AE%97%E7%9A%84%E5%95%8A%EF%BC%9F)
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求通项:a1=3,且a(n+1)=a²n(n∈正整数)答案说用迭代法求~迭代法是怎么算的啊?
求通项:a1=3,且a(n+1)=a²n(n∈正整数)
答案说用迭代法求~迭代法是怎么算的啊?
求通项:a1=3,且a(n+1)=a²n(n∈正整数)答案说用迭代法求~迭代法是怎么算的啊?
a(n+1)=an²
=a(n-1)⁴
=...
=a1^(2ⁿ)
=3^(2ⁿ)
an=3^[2^(n-1)]
n=1时,a1=3^(2^0)=3,同样满足通项公式
数列{an}的通项公式为an=3^[2^(n-1)]
或者这样
a1=3>0
假设当n=k(k∈N+)时,ak>0,则a(k+1)=ak²>0
k为任意正整数,因此对于任意正整数n,an恒>0
a(n+1)=an²
log3[a(n+1)]=log3(an²)=2log3(an)
log3(a1)=log3(3)=1
log3[a(n+1)]/log3(an)=2,为定值
数列{log3(an)}是以1为首项,2为公比的等比数列
log3(an)=1×2^(n-1)=2^(n-1)
an=3^[2^(n-1)]
n=1时,a1=3^(2^0)=3,同样满足通项公式
数列{an}的通项公式为an=3^[2^(n-1)]
答:
A1=3=3^1
A(n+1)=(An)^2
所以:
A2=(A1)^2=3^2
A3=(A2)^2=3^4
A4=(A3)^2=3^8
......
An=3^[ 2^(n-1)]
A(n+1)=3^[2^n]={ 3^[2^(n-1)] }^2=[ An ]^2,符合题目条件
所以:An=3^[ 2^(n-1)]