已知函数f(x)=2^x-2^(-x),数列{an}满足f(log2 an)=-2n.(1)求数列{an}的通项公式.f(log2 an) = -2n=> 2^(log2 an)-2^(-(log2 an)) = -2n=> an - 1/an = -2n=> an^2 +2*n*an -1 = 0=> an = -n+sqrt(n^2+1) 或 an = -n-sqrt(n^2+1)由于题目中有 logx an,所以an>0,所以只能取前一个an = -n+sqrt(n^2+1)为什么2^(log2 an)-2^(-(log2 an)) = -2n能得到 an - 1/an = -2n
问题描述：
已知函数f(x)=2^x-2^(-x),数列{an}满足f(log2 an)=-2n.(1)求数列{an}的通项公式.
f(log2 an) = -2n
=> 2^(log2 an)-2^(-(log2 an)) = -2n
=> an - 1/an = -2n
=> an^2 +2*n*an -1 = 0
=> an = -n+sqrt(n^2+1) 或 an = -n-sqrt(n^2+1)
由于题目中有 logx an,所以an>0,所以只能取前一个
an = -n+sqrt(n^2+1)
为什么2^(log2 an)-2^(-(log2 an)) = -2n能得到 an - 1/an = -2n
答
f(x)=2^x-2^(-x)
f(logan) = -2n
2^[logan] - 2^[-logan] = -2n
2^[logan] - 2^[log(1/an)] = -2n
an -1/an = -2n
(an)^2 +2nan -1 =0
an = -n+√(n^2+1)
答
你那一步就是用到以下这两个常用公式呀
a^(log a n) =a
-(log a n) = log a (1/n)