Answer
$t_1 =-\dfrac{1}{6}
\\t_2= \dfrac{1}{12}
\\t_3=-\dfrac{1}{20}
\\t_4= \dfrac{1}{30}
\\t_5=-\dfrac{1}{42}$
Work Step by Step
We are given that {$t_n$} $=\dfrac{(-1)^n}{(n+1)(n+2)}$
In order to determine the first five terms, we will have to substitute $n=2,3,4,5$ into the given sequence {$t_n$}:
$t_1 = \dfrac{(-1)^1}{(1+1)(1+2)}=\dfrac{-1}{(2)(3)}=-\dfrac{1}{6}
\\t_2= \dfrac{(-1)^2}{(2+1)(2+2)}=\dfrac{1}{(3)(4)}=\dfrac{1}{12}
\\t_3= \dfrac{(-1)^3}{(3+1)(3+2)}=\dfrac{-1}{(4)(5)}=-\dfrac{1}{20}
\\t_4= \dfrac{(-1)^4}{(4+1)(4+2)}=\dfrac{1}{(5)(6)}=\dfrac{1}{30}
\\t_5= \dfrac{(-1)^5}{(5+1)(5+2)}=\dfrac{-1}{(6)(7)}=-\dfrac{1}{42}$
