1
Ejemplo resuelto de teorema del binomio
$\left(x+3\right)^5$
2
Podemos expandir la expresión $\left(x+3\right)^5$ usando el binomio de Newton, el cual es una fórmula que nos permite obtener la forma expandida de un binomio elevado a un número entero $n$. La fórmula tal cual es: $\displaystyle(a\pm b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)a^{n-k}b^k=\left(\begin{matrix}n\\0\end{matrix}\right)a^n\pm\left(\begin{matrix}n\\1\end{matrix}\right)a^{n-1}b+\left(\begin{matrix}n\\2\end{matrix}\right)a^{n-2}b^2\pm\dots\pm\left(\begin{matrix}n\\n\end{matrix}\right)b^n$. El número de términos que resultan de la expansión es siempre igual a $n+1$. Los coeficientes $\left(\begin{matrix}n\\k\end{matrix}\right)$ son números combinatorios los cuales corresponden a la fila enésima del triángulo de Tartaglia (o triángulo de Pascal). En la fórmula, podemos observar que el exponente de $a$ va disminuyendo, de $n$ a $0$, mientras que el exponente de $b$ va aumentando, de $0$ a $n$. Si uno de los términos del binomio es negativo, se alternan los signos positivos y negativos.
$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}3^{0}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}3^{2}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
3
Calcular la potencia $3^{0}$
$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}3^{2}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
4
Calcular la potencia $3^{2}$
$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
5
Calcular la potencia $3^{3}$
$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
6
Calcular la potencia $3^{4}$
$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
7
Calcular la potencia $3^{5}$
$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
8
Cualquier expresión elevada a la potencia uno es igual a esa misma expresión
$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
9
Cualquier expresión algebraica multiplicada por uno es igual a esa misma expresión
$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
10
Cualquier expresión matemática elevada a la potencia $0$ es igual a $1$
$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 1\cdot 243$
11
Multiplicar $1$ por $243$
$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+243\cdot \left(\begin{matrix}5\\5\end{matrix}\right)$
12
Calcular el coeficiente binomial $\left(\begin{matrix}5\\0\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$x^{5}\frac{5!}{\left(0!\right)\left(5!\right)}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+243\cdot \left(\begin{matrix}5\\5\end{matrix}\right)$
13
Calcular el coeficiente binomial $\left(\begin{matrix}5\\1\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$x^{5}\frac{5!}{\left(0!\right)\left(5!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+243\cdot \left(\begin{matrix}5\\5\end{matrix}\right)$
14
Calcular el coeficiente binomial $\left(\begin{matrix}5\\2\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$x^{5}\frac{5!}{\left(0!\right)\left(5!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+243\cdot \left(\begin{matrix}5\\5\end{matrix}\right)$
15
Calcular el coeficiente binomial $\left(\begin{matrix}5\\3\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$x^{5}\frac{5!}{\left(0!\right)\left(5!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+243\cdot \left(\begin{matrix}5\\5\end{matrix}\right)$
16
Calcular el coeficiente binomial $\left(\begin{matrix}5\\4\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$x^{5}\frac{5!}{\left(0!\right)\left(5!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x\frac{5!}{\left(4!\right)\left(1!\right)}+243\cdot \left(\begin{matrix}5\\5\end{matrix}\right)$
17
Calcular el coeficiente binomial $\left(\begin{matrix}5\\5\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$x^{5}\frac{5!}{\left(0!\right)\left(5!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x\frac{5!}{\left(4!\right)\left(1!\right)}+243\left(\frac{5!}{\left(5!\right)\left(0!\right)}\right)$
18
Simplificar la fracción $\frac{5!}{\left(0!\right)\left(5!\right)}$ por $5!$
$x^{5}\frac{1}{0!}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x\frac{5!}{\left(4!\right)\left(1!\right)}+243\left(\frac{5!}{\left(5!\right)\left(0!\right)}\right)$
19
Simplificar la fracción $\frac{5!}{\left(5!\right)\left(0!\right)}$ por $5!$
$x^{5}\frac{1}{0!}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x\frac{5!}{\left(4!\right)\left(1!\right)}+243\left(\frac{1}{0!}\right)$
20
Multiplicar la fracción por el término
$\frac{x^{5}}{0!}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x\frac{5!}{\left(4!\right)\left(1!\right)}+243\left(\frac{1}{0!}\right)$
21
Multiplicar la fracción por el término
$\frac{x^{5}}{0!}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x\frac{5!}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
22
Multiplicando la fracción por el término $3x^{4}$
$\frac{x^{5}}{0!}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x\frac{5!}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
23
Multiplicando la fracción por el término $9x^{3}$
$\frac{x^{5}}{0!}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x\frac{5!}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
24
Multiplicando la fracción por el término $27x^{2}$
$\frac{x^{5}}{0!}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+81x\frac{5!}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
25
Multiplicando la fracción por el término $81x$
$\frac{x^{5}}{0!}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
26
El factorial de $0$ es $1$
$\frac{x^{5}}{1}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
27
El factorial de $1$ es $1$
$\frac{x^{5}}{1}+\frac{3\left(5!\right)\left(x^{4}\right)}{1\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
28
El factorial de $4$ es $24$
$\frac{x^{5}}{1}+\frac{3\left(5!\right)\left(x^{4}\right)}{1\cdot 24}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
29
El factorial de $5$ es $120$
$\frac{x^{5}}{1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
30
El factorial de $2$ es $2$
$\frac{x^{5}}{1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{9\left(5!\right)\left(x^{3}\right)}{2\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
31
El factorial de $3$ es $6$
$\frac{x^{5}}{1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{9\left(5!\right)\left(x^{3}\right)}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
32
El factorial de $5$ es $120$
$\frac{x^{5}}{1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
33
El factorial de $3$ es $6$
$\frac{x^{5}}{1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{6\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
34
El factorial de $2$ es $2$
$\frac{x^{5}}{1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
35
El factorial de $5$ es $120$
$\frac{x^{5}}{1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
36
Multiplicar $1$ por $24$
$\frac{x^{5}}{1}+\frac{120\cdot 3x^{4}}{24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
37
Multiplicar $120$ por $3$
$\frac{x^{5}}{1}+\frac{360x^{4}}{24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
38
Multiplicar $2$ por $6$
$\frac{x^{5}}{1}+\frac{360x^{4}}{24}+\frac{120\cdot 9x^{3}}{12}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
39
Multiplicar $120$ por $9$
$\frac{x^{5}}{1}+\frac{360x^{4}}{24}+\frac{1080x^{3}}{12}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
40
Multiplicar $6$ por $2$
$\frac{x^{5}}{1}+\frac{360x^{4}}{24}+\frac{1080x^{3}}{12}+\frac{120\cdot 27x^{2}}{12}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
41
Multiplicar $120$ por $27$
$\frac{x^{5}}{1}+\frac{360x^{4}}{24}+\frac{1080x^{3}}{12}+\frac{3240x^{2}}{12}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
42
Cualquier expresión matemática dividida por uno ($1$) es igual a esa misma expresión
$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
43
El factorial de $4$ es $24$
$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{81x\left(5!\right)}{24\left(1!\right)}+\frac{243}{0!}$
44
El factorial de $1$ es $1$
$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{81x\left(5!\right)}{24\cdot 1}+\frac{243}{0!}$
45
El factorial de $5$ es $120$
$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{120\cdot 81x}{24\cdot 1}+\frac{243}{0!}$
46
El factorial de $0$ es $1$
$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{120\cdot 81x}{24\cdot 1}+\frac{243}{1}$
47
Multiplicar $24$ por $1$
$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{120\cdot 81x}{24}+\frac{243}{1}$
48
Multiplicar $120$ por $81$
$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{9720x}{24}+\frac{243}{1}$
49
Dividir $243$ entre $1$
$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{9720x}{24}+243$
50
Sacar el $\frac{9720}{24}$ de la fracción
$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$
Respuesta Final
$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$