Answer
The first five terms are:
$t_1 = -\dfrac{1}{6};
\\t_2= \dfrac{1}{12};
\\t_3= -\dfrac{1}{20};
\\t_4= \dfrac{1}{30};
\\t_5= -\dfrac{1}{42}$
Work Step by Step
Substitute $n=1, 2, 3, 4, 5$ to obtain:
$t_1 = \dfrac{(-1)^1}{(1+1)(1+2)}=\dfrac{-1}{2(3)}=-\dfrac{1}{6};
\\t_2= \dfrac{(-1)^2}{(2+1)(2+2)}=\dfrac{1}{3(4)}=\dfrac{1}{12};
\\t_3= \dfrac{(-1)^3}{(3+1)(3+2)}=\dfrac{-1}{4(5)}=-\dfrac{1}{20};
\\t_4= \dfrac{(-1)^4}{(4+1)(4+2)}=\dfrac{1}{5(6)}=\dfrac{1}{30};
\\t_5= \dfrac{(-1)^5}{(5+1)(5+2)}=\dfrac{-1}{6(7)}=-\dfrac{1}{42}$
