What is 2/n ((n!)/(2^{n+1)}) ?

The solution to 2/n ((n!)/(2^{n+1)}) is (n!)/(n*2^n)

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simplify 2/n ((n!)/(2^{n+1)})

simplify

\frac{2}{n} (\frac{n !}{2 ^{n + 1}})

\frac{n !}{n \cdot 2 ^{n}}

Solution steps

\frac{2}{n} (\frac{n !}{2 ^{n + 1}})

Apply rule:

(a) = a

(\frac{n !}{2 ^{n + 1}}) = \frac{n !}{2 ^{n + 1}}

= \frac{2}{n} \cdot \frac{n !}{2 ^{n + 1}}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{2 n !}{n \cdot 2 ^{n + 1}}

Simplify

\frac{2}{2 ^{n + 1}} : \frac{1}{2 ^{n}}

= \frac{n !}{n \cdot 2 ^{n}}

8.5867 (2.0409 6^{3}) - 6.44 (2.04096) + 4 = x

x + 2 \leq \frac{1}{x + 2}

- 7200 (1 + x)^{2} + 5300 (1 + x) + 8000 = 0

27 a^{3} - b^{6}

4 (y + 1) - y = 3 (y - 1) + 3

What is 2/n ((n!)/(2^{n+1)}) ?
The solution to 2/n ((n!)/(2^{n+1)}) is (n!)/(n*2^n)