Desarrolla $$$\left(s - 5 v\right)^{5}$$$

Desarrolla $$$\left(s - 5 v\right)^{5}$$$.

La expansión viene dada por la siguiente fórmula: $$$\left(a + b\right)^{n} = \sum_{k=0}^{n} {\binom{n}{k}} a^{n - k} b^{k}$$$, donde $$${\binom{n}{k}} = \frac{n!}{\left(n - k\right)! k!}$$$ y $$$n! = 1 \cdot 2 \cdot \ldots \cdot n$$$.

Tenemos que $$$a = s$$$, $$$b = - 5 v$$$ y $$$n = 5$$$.

Por lo tanto, $$$\left(s - 5 v\right)^{5} = \sum_{k=0}^{5} {\binom{5}{k}} s^{5 - k} \left(- 5 v\right)^{k}$$$.

Ahora, calcula el producto para cada valor de $$$k$$$ desde $$$0$$$ hasta $$$5$$$.

$$$k = 0$$$: $$${\binom{5}{0}} s^{5 - 0} \left(- 5 v\right)^{0} = \frac{5!}{\left(5 - 0\right)! 0!} s^{5 - 0} \left(- 5 v\right)^{0} = s^{5}$$$

$$$k = 1$$$: $$${\binom{5}{1}} s^{5 - 1} \left(- 5 v\right)^{1} = \frac{5!}{\left(5 - 1\right)! 1!} s^{5 - 1} \left(- 5 v\right)^{1} = - 25 s^{4} v$$$

$$$k = 2$$$: $$${\binom{5}{2}} s^{5 - 2} \left(- 5 v\right)^{2} = \frac{5!}{\left(5 - 2\right)! 2!} s^{5 - 2} \left(- 5 v\right)^{2} = 250 s^{3} v^{2}$$$

$$$k = 3$$$: $$${\binom{5}{3}} s^{5 - 3} \left(- 5 v\right)^{3} = \frac{5!}{\left(5 - 3\right)! 3!} s^{5 - 3} \left(- 5 v\right)^{3} = - 1250 s^{2} v^{3}$$$

$$$k = 4$$$: $$${\binom{5}{4}} s^{5 - 4} \left(- 5 v\right)^{4} = \frac{5!}{\left(5 - 4\right)! 4!} s^{5 - 4} \left(- 5 v\right)^{4} = 3125 s v^{4}$$$

$$$k = 5$$$: $$${\binom{5}{5}} s^{5 - 5} \left(- 5 v\right)^{5} = \frac{5!}{\left(5 - 5\right)! 5!} s^{5 - 5} \left(- 5 v\right)^{5} = - 3125 v^{5}$$$

Por lo tanto, $$$\left(s - 5 v\right)^{5} = s^{5} - 25 s^{4} v + 250 s^{3} v^{2} - 1250 s^{2} v^{3} + 3125 s v^{4} - 3125 v^{5}.$$$

$$$\left(s - 5 v\right)^{5} = s^{5} - 25 s^{4} v + 250 s^{3} v^{2} - 1250 s^{2} v^{3} + 3125 s v^{4} - 3125 v^{5}$$$A