> ## Documentation Index
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> Use this file to discover all available pages before exploring further.

# Binomial Coefficients and Identities

## The Binomial Theorem

The binomial theorem gives the coefficients of the expansion of powers of binomial expressions

> Teorema binomial memberikan koefisien ekspansi pangkat ekspresi binomial

A **binomial** expression is simply the sum of two terms, such as $x + y$.

> Ekspresi **binomial** merupakan jumlah dari dua bentuk, seperti $x + y$.

### Example 1

<Card>
  The expansion of $(x + y)^3$ can be found using combinatorial reasoning
  instead of multiplying the three terms out.

  <br />

  When $(x + y)^3 = (x + y)(x + y)(x + y)$ is expanded, all products of a term
  in the first sum, a term in the second sum, and a term in the third sum are
  added.

  <br />

  Terms of the form $x^3~, x^2y~, xy^2$, and $y^3$ arise.

  <br />

  To obtain a term of the form $x^3$, an x must be chosen in each of the sums,
  and this can be done in only one way. Thus, the $x^3$ term in the product has
  a coefficient of 1

  <br />

  It continues for $x^2y, xy^2$, and $y^3$ and it follows that

  <br />

  ![Binomial
  Coefficients](https://i.ibb.co/jb7zJDy/Screenshot-2024-04-12-220006.png)

  <br />

  Let x and y be variables, and let n be a nonnegative integer. Then

  <br />

  ![Binomial](https://i.ibb.co/Hr0zFCf/Screenshot-2024-04-12-220306.png)
</Card>

### Example 2

<Card>
  What is the expansion of $(x + y)^4$?

  **Solution:**

  <Check>
    From the binomial theorem it follows that

    <br />

    $(x + y)^4 = \sum_{j=0}^{4} \binom{4}{j} x^{4-j} y^j$

    <br />

    $~~~~~~~~~~~~~~~= \binom{4}{0} x^4 + \binom{4}{1} x^3 y + \binom{4}{2} x^2 y^2 + \binom{4}{3} x y^3 + \binom{4}{4} y^4$

    <br />

    $~~~~~~~~~~~~~~~= x^4 + 4x^3 y + 6x^2 y^2 + 4x y^3 + y^4$
  </Check>
</Card>

### Example 3

<Card>
  What is the coefficient of $x^{12} y^{13}$ in the expansion of $(x + y)^25$?

  **Solution:**

  <Check>
    From the binomial theorem it follows that this coefficient is

    <br />

    $\binom{25}{13} = \dfrac{25!}{13!~12!} = 5~200~300$
  </Check>
</Card>

### Corollary 1

<Card>
  Let $n$ be a nonnegative integer. Then

  <Check>
    $\sum_{k = 0}^{n} \binom{n}{k} = 2^n$
  </Check>
</Card>