Question

# Substitute n=1, 2, 3, 4, 5 and write the first five terms of the sequence \left\{\frac{(-1)^{n-1}x^{2n-1}}{(2n-1)!}\right\}

Sequences
Substitute n=1, 2, 3, 4, 5 and write the first five terms of the sequence
$$\left\{\frac{(-1)^{n-1}x^{2n-1}}{(2n-1)!}\right\}$$

$$\left\{\frac{(-1)^{n-1}x^{2n-1}}{(2n-1)!}\right\}$$
$$\left\{\frac{(-1)^{1-1}x^{2(1)-1}}{(2(1)-1)!}\right\}=\frac{(-1)^{0}x^{1}}{(1)!}=x$$
$$\left\{\frac{(-1)^{2-1}x^{2(2)-1}}{(2(2)-1)!}\right\}=\frac{(-1)^{1}x^{3}}{(3)!}=-\frac{x^{3}}{6}$$
$$\left\{\frac{(-1)^{3-1}x^{2(3)-1}}{(2(3)-1)!}\right\}=\frac{(-1)^{2}x^{5}}{(5)!}=\frac{x^{5}}{120}$$
$$\left\{\frac{(-1)^{4-1}x^{2(4)-1}}{(2(4)-1)!}\right\}=\frac{(-1)^{3}x^{7}}{(7)!}=-\frac{x^{7}}{5040}$$
$$\left\{\frac{(-1)^{5-1}x^{2(5)-1}}{(2(5)-1)!}\right\}=\frac{(-1)^{4}x^{9}}{(9)!}=\frac{x^{9}}{362880}$$