Answer
$-\dfrac{1}{6},\dfrac{1}{12},-\dfrac{1}{20},\dfrac{1}{30},-\dfrac{1}{42}$
Work Step by Step
We are given the sequence:
$\{t_n\}=\left\{\dfrac{(-1)^n}{(n+1)(n+2)}\right\}$
Determine the first 5 terms of the sequence by substituting 1, 2, 3, 4, 5 for $n$:
$t_1=\dfrac{(-1)^n}{(n+1)(n+2)}=\dfrac{-1}{2\cdot 3}=-\dfrac{1}{6}$
$t_2=\dfrac{(-1)^n}{(n+1)(n+2)}=\dfrac{1}{3\cdot 4}=\dfrac{1}{12}$
$t_3=\dfrac{(-1)^n}{(n+1)(n+2)}=\dfrac{-1}{4\cdot 5}=-\dfrac{1}{20}$
$t_4=\dfrac{(-1)^n}{(n+1)(n+2)}=\dfrac{1}{5\cdot 6}=\dfrac{1}{30}$
$t_5=\dfrac{(-1)^n}{(n+1)(n+2)}=\dfrac{-1}{6\cdot 7}=-\dfrac{1}{42}$
