Question

Aplicando los números combinatorios, desarrolla:

Answer 1

Antes de empezar debemos conocer la combinatoria y su formula, porque la desarrollaremos para resolverla.
${m \choose n}= \frac{m!}{n!(m-n)!}$

Aplicando los números combinatorios, desarrolla:
a) (4x³-3y)⁴=
Haremos uso de la combinatoria, iremos incrementando la potencia del primer elemento y reduciendo la del segundo.

$(4x^3-3y)^4={4 \choose 0}(4x^3)^4+{4 \choose 1}(4x^3)^3(-3y)+{4 \choose 2}(4x^3)^2(-3y)^2 \\ \hspace*{20mm} +{4 \choose 3}(4x^3)(-3y)^3+{4 \choose 4}(-3y)^4 \\ \\ =(1)(256x^{12})+(4)(64x^{9})(-3y)+(6)(16x^6)(9y^2) \\ \hspace*{20mm} +(4)(4x^3)(-27y^3)+(1)(81y^4) \\ \\ =256x^{12}-768x^9y+864x^6y^2-432x^3y^3+81y^4$


b) $(\frac{-2}{x^2}+4x)^7$

Resolvemos de modo similar
$( \frac{-2}{x^2} +4x)^7={7 \choose 0}(\frac{-2}{x^2})^7+{7 \choose 1}(\frac{-2}{x^2})^6(4x)+{7 \choose 2}(\frac{-2}{x^2})^5(4x)^2 \\ \\ \hspace*{20mm} {7 \choose 3}(\frac{-2}{x^2})^4(4x)^3+{7 \choose 4}(\frac{-2}{x^2})^3(4x)^4+{7 \choose 5}(\frac{-2}{x^2})^2(4x)^5 \\ \\ \hspace*{20mm} {7 \choose 6}(\frac{-2}{x^2})(4x)^6+{7 \choose 7}(4x)^7$

$\\ \hspace*{20mm} =(1)(\frac{-128}{x^{14}})+(7)(\frac{64}{x^{12}})(4x)+(21)(\frac{-32}{x^{10}})16x^2)+(35)( \frac{16}{x^8} )(64x^3)\\ \\ \hspace*{20mm} +(35)(\frac{-8}{x^6})(256x^4)+(21)(\frac{4}{x^4})(1024x^5)+(7)(\frac{-2}{x^2})(4096x^6)+16384x^7\\ \\ \hspace*{20mm} =-\frac{128}{x^{14}}+\frac{-1792}{x^{11}}-\frac{10752}{x^8}+\frac{35840}{x^5}-\frac{71680}{x^2}+86016x \\ \\ \hspace*{20mm} -57344x^4+16384x^7$