Answer
$-\frac{1}{2},\frac{2}{3},-\frac{3}{7},\frac{4}{25},-\frac{5}{121}$
Work Step by Step
To find the first five terms of the sequence $a_n = \frac{(-1)^{(n)} n}{n!+1}$, we must plug in $n=1, n=2, n=3, n=4,$ and $n=5.$
$a_1 = \frac{(-1)^{1}\times 1}{1!+1}=\frac{-1}{1+1}=-\frac{1}{2}$
$a_2 =\frac{(-1)^{2} \times 2}{2!+1}=\frac{2}{2+1}=\frac{2}{3}$
$a_3 = \frac{(-1)^{3} \times 3}{3!+1}=-\frac{3}{6+1}=-\frac{3}{7}$
$a_4 =\frac{(-1)^{4} \times 4}{4!+1}=\frac{4}{24+1}=\frac{4}{25}$
$a_5 = \frac{(-1)^{5} \times 5}{5!+1}=\frac{-5}{120+1}=-\frac{5}{121}$
Hence we see that the first five terms are $-\frac{1}{2},\frac{2}{3},-\frac{3}{7},\frac{4}{25},-\frac{5}{121}$
