已知函數(shù)f(x)=2^x-2^(-x),數(shù)列{an}滿足f(log2 an)=-2n.(1)求數(shù)列{an}的通項(xiàng)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)
由于題目中有 logx an,所以an>0,所以只能取前一個(gè)
an = -n+sqrt(n^2+1)
為什么2^(log2 an)-2^(-(log2 an)) = -2n能得到 an - 1/an = -2n