Answer:
[tex]f(n)=(-1)^{n-1}\cdot\frac{n}{n+5}[/tex]Explanation:
The sequence is formed by fractions where the sign is changing from term to term. Then, it will have a power of (-1). Additionally, the numerator of the fractions starts at 1 and the denominator of the fraction starts at 6.
Then, the nth term will be
[tex]f(n)=(-1)^{n-1}\cdot\frac{n}{n+5}[/tex]Replacing n by 1, 2, 3, and 4, we get:
[tex]\begin{gathered} f(1)=(-1)^{1-1}\cdot\frac{1}{1+5}=(-1)^0\cdot\frac{1}{6}=\frac{1}{6} \\ f(2)=(-1)^{2-1}\cdot\frac{2}{2+5}=(-1)^1\cdot\frac{2}{7}=\frac{-2}{7} \\ f(3)=(-1)^{3-1}\cdot\frac{3}{3+5}=(-1)^2\cdot\frac{3}{8}=\frac{3}{8} \\ f(4)=(-1)^{4-1}\cdot\frac{4}{4+5}=(-1)^3\cdot\frac{4}{9}=\frac{-4}{9} \end{gathered}[/tex]Therefore, the answer is
[tex]f(n)=(-1)^{n-1}\cdot\frac{n}{n+5}[/tex]