Answer
$1,\dfrac{-1}{3},\dfrac{1}{7},\dfrac{-1}{15}$
Work Step by Step
Need to find the first four terms of $a_n=\dfrac{(-1)^{n+1}}{2^n-1}$ when $n=1,2,3,4$
Thus,
$a_1=\dfrac{(-1)^{1+1}}{2^1-1}=1$
$a_2=\dfrac{(-1)^{2+1}}{2^2-1}=\dfrac{-1}{3}$
$a_3=\dfrac{(-1)^{3+1}}{2^3-1}=\dfrac{1}{7}$
$a_4=\dfrac{(-1)^{4+1}}{2^4-1}=\dfrac{-1}{15}$
Hence, the first four terms are: $1,\dfrac{-1}{3},\dfrac{1}{7},\dfrac{-1}{15}$
