# Substitute n=1, 2, 3, 4, 5 to find the first five sequences in the given sequence \left\{\frac{(-1)^{n-1}}{2\cdot4\cdot6\dotsm2n}\right\}

Substitute n=1, 2, 3, 4, 5 to find the first five sequences in the given sequence
$\left\{\frac{\left(-1{\right)}^{n-1}}{2\cdot 4\cdot 6\cdots 2n}\right\}$
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$\left\{\frac{\left(-1{\right)}^{n-1}}{2\cdot 4\cdot 6\cdots 2n}\right\}$
$\left\{\frac{\left(-1{\right)}^{1-1}}{2\cdot 4\cdot 6\cdots 2\left(1\right)}\right\}=\frac{\left(-1{\right)}^{0}}{2}=\frac{1}{2}$
$\left\{\frac{\left(-1{\right)}^{2-1}}{2\cdot 4\cdot 6\cdots 2\left(2\right)}\right\}=\frac{\left(-1{\right)}^{1}}{2\cdot 4}=-\frac{1}{8}$
$\left\{\frac{\left(-1{\right)}^{3-1}}{2\cdot 4\cdot 6\cdots 2\left(3\right)}\right\}=\frac{\left(-1{\right)}^{2}}{2\cdot 4\cdot 6}=\frac{1}{48}$
$\left\{\frac{\left(-1{\right)}^{4-1}}{2\cdot 4\cdot 6\cdots 2\left(4\right)}\right\}=\frac{\left(-1{\right)}^{3}}{2\cdot 4\cdot 6\cdot 8}=-\frac{1}{384}$
$\left\{\frac{\left(-1{\right)}^{5-1}}{2\cdot 4\cdot 6\cdots 2\left(5\right)}\right\}=\frac{\left(-1{\right)}^{4}}{2\cdot 4\cdot 6\cdot 8\cdot 10}=\frac{1}{3840}$