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\documentclass{article}
\usepackage{graphicx} % Required for inserting images
\title{Laboratorio}
\author{Alexx}
\date{January 2024}
\begin{document}
\maketitle
\textbf{Encuentre el término de orden $r=4$ $(x+4)^4$}
\[
\frac{n(n-1)\ldots(n-r+2)}{(r-1)!}a^{n-r+1}b^{r-1}
\]
\[
= \frac{4(4-1)(4-2)}{(4-1)!}x^{4-4+1}(4^{4-1})
\]
\[
= \frac{4(3)(2)}{(3)!}x^{4-4+1}(4^{4-1})
\]
\[
= \frac{24}{(3)!}x^{4-4+1}(4^{4-1})
\]
\[
= \frac{24}{6}x^{4-4+1}(4^{4-1})
\]
\[
= 4x^{4-4+1}(4^{4-1}) = (4x^{1})(4^{3}) = 256x
\]
\textbf{Desarrolle $(3x+5)^4$}
\[
(3x+5)^4
\]
\[
= 81x^4+4(3x)^{4-1}(5)+\frac{4(4-1)}{2!}(3x)^{4-2}(5^2)+\frac{4(4-1)(4-2)}{3!}(3x)^{4-3}(5^3)+5^4
\]
\[
= 81x^4+4(27x)^{3}(5)+\frac{4(4-1)}{2!}(3x)^{4-2}(5^2)+\frac{4(4-1)(4-2)}{3!}(3x)^{4-3}(5^3)+5^4
\]
\[
= 81x^4+540x^3+\frac{4(3)}{2}(3x)^{4-2}(5^2)+\frac{4(4-1)(4-2)}{3!}(3x)^{4-3}(5^3)+5^4
\]
\[
= 81x^4+540x^3+(6)(9x^2)(5^2)+\frac{4(4-1)(4-2)}{3!}(3x)^{4-3}(5^3)+5^4
\]
\[
= 81x^4+540x^3+1,350x^2+\frac{4(3)(2)}{6}(3x)^{4-3}(5^3)+5^4
\]
\[
= 81x^4+540x^3+1,350x^2+(4)(3x)^{4-3}(9)+5^4
\]
\[
= 81x^4+540x^3+1,350x^2+(12x)(5^3)+5^4
\]
\[
= 81x^4+540x^3+1,350x^2+(12x)(125)+5^4
\]
\[
= 81x^4+540x^3+1,350x^2+1500x+5^4
\]
\[
= 81x^4+540x^3+1,350x^2+1500x+625
\]
\end{document}
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