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1

Solved example of binomial theorem

$\left(x+3\right)^5$
2

We can expand the expression $\left(x+3\right)^5$ using Newton's binomial theorem, which is a formula that allow us to find the expanded form of a binomial raised to a positive integer $n$. The formula is as follows: $\displaystyle(a\pm b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)a^{n-k}b^k=\left(\begin{matrix}n\\0\end{matrix}\right)a^n\pm\left(\begin{matrix}n\\1\end{matrix}\right)a^{n-1}b+\left(\begin{matrix}n\\2\end{matrix}\right)a^{n-2}b^2\pm\dots\pm\left(\begin{matrix}n\\n\end{matrix}\right)b^n$
The number of terms resulting from the expansion always equals $n + 1$. The coefficients $\left(\begin{matrix}n\\k\end{matrix}\right)$ are combinatorial numbers which correspond to the nth row of the Tartaglia triangle (or Pascal's triangle). In the formula, we can observe that the exponent of $a$ decreases, from $n$ to $0$, while the exponent of $b$ increases, from $0$ to $n$. If one of the binomial terms is negative, the positive and negative signs alternate.

$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}3^{0}+\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}3^{1}+\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}3^{2}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
3

Calculate the power $3^{0}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}3^{1}+\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}3^{2}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$

$\frac{\left(5!\right)\left(x^{5}\right)}{\left(0!\right)\left(5!\right)}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

Simplify the fraction by $5!$

$\frac{x^{5}}{0!}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $0$ is $1$

$\frac{x^{5}}{1}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

Any expression divided by one ($1$) is equal to that same expression

$x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
4

Calculate the binomial coefficient $\left(\begin{matrix}5\\0\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

$x^{5}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $5$ is $120$

$x^{5}+\frac{120\cdot 3x^{4}}{\left(1!\right)\left(4!\right)}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $1$ is $1$

$x^{5}+\frac{120\cdot 3x^{4}}{1\left(4!\right)}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $4$ is $24$

$x^{5}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

Any expression multiplied by $1$ is equal to itself

$x^{5}+\frac{360x^{4}}{24}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
5

Calculate the binomial coefficient $\left(\begin{matrix}5\\1\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$x^{5}+\frac{360x^{4}}{24}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
6

Take $\frac{360}{24}$ out of the fraction

$x^{5}+15x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

$x^{5}+15x^{4}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $5$ is $120$

$x^{5}+15x^{4}+\frac{120\cdot 9x^{3}}{\left(2!\right)\left(3!\right)}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $2$ is $2$

$x^{5}+15x^{4}+\frac{120\cdot 9x^{3}}{2\left(3!\right)}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $3$ is $6$

$x^{5}+15x^{4}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
7

Calculate the binomial coefficient $\left(\begin{matrix}5\\2\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$x^{5}+15x^{4}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
8

Take $\frac{120}{2}$ out of the fraction

$x^{5}+15x^{4}+\frac{540x^{3}}{6}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

$x^{5}+15x^{4}+\frac{540x^{3}}{6}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $5$ is $120$

$x^{5}+15x^{4}+\frac{540x^{3}}{6}+\frac{120\cdot 27x^{2}}{\left(3!\right)\left(2!\right)}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $3$ is $6$

$x^{5}+15x^{4}+\frac{540x^{3}}{6}+\frac{120\cdot 27x^{2}}{6\left(2!\right)}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $2$ is $2$

$x^{5}+15x^{4}+\frac{540x^{3}}{6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
9

Calculate the binomial coefficient $\left(\begin{matrix}5\\3\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$x^{5}+15x^{4}+\frac{540x^{3}}{6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
10

Take $\frac{540}{6}$ out of the fraction

$x^{5}+15x^{4}+90x^{3}+\frac{3240x^{2}}{12}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

$x^{5}+15x^{4}+90x^{3}+\frac{3240x^{2}}{12}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $5$ is $120$

$x^{5}+15x^{4}+90x^{3}+\frac{3240x^{2}}{12}+\frac{120\cdot 81x}{\left(4!\right)\left(1!\right)}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $4$ is $24$

$x^{5}+15x^{4}+90x^{3}+\frac{3240x^{2}}{12}+\frac{120\cdot 81x}{24\left(1!\right)}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

The factorial of $1$ is $1$

$x^{5}+15x^{4}+90x^{3}+\frac{3240x^{2}}{12}+\frac{120\cdot 81x}{24\cdot 1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

Any expression multiplied by $1$ is equal to itself

$x^{5}+15x^{4}+90x^{3}+\frac{3240x^{2}}{12}+\frac{9720x}{24}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
11

Calculate the binomial coefficient $\left(\begin{matrix}5\\4\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$x^{5}+15x^{4}+90x^{3}+\frac{3240x^{2}}{12}+\frac{9720x}{24}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$
12

Take $\frac{3240}{12}$ out of the fraction

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{9720x}{24}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243$

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{9720x}{24}+\frac{243\left(5!\right)}{\left(5!\right)\left(0!\right)}$

Simplify the fraction by $5!$

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{9720x}{24}+\frac{243}{0!}$

The factorial of $0$ is $1$

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{9720x}{24}+\frac{243}{1}$

Any expression divided by one ($1$) is equal to that same expression

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{9720x}{24}+243$
13

Calculate the binomial coefficient $\left(\begin{matrix}5\\5\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{9720x}{24}+243$
14

Take $\frac{9720}{24}$ out of the fraction

$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$

Final Answer

$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$

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