Teach me binomial theorem into details

The general formula for the Binomial Theorem is:

$$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

Here's a breakdown of each component:

• $n$: This is the power to which the binomial $(a+b)$ is raised. It must be a non-negative integer.
• $a$ and $b$: These are the two terms of the binomial.
• $k$: This is an index that ranges from 0 to $n$. Each value of $k$ corresponds to a specific term in the expansion.
• $\binom{n}{k}$: This is the binomial coefficient, read as "n choose k". It represents the number of ways to choose $k$ items from a set of $n$ items.

The binomial coefficient $\binom{n}{k}$ is calculated using factorials:

$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

where $n!$ (n factorial) is the product of all positive integers up to $n$ (e.g., $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$). By definition, $0! = 1$.

Let's expand $(x+y)^3$ using the Binomial Theorem as an example:
Here, $n=3$, $a=x$, and $b=y$. We will sum terms for $k=0, 1, 2, 3$.

For $k=0$:

$$\binom{3}{0} x^{3-0} y^0 = \frac{3!}{0!(3-0)!} x^3 y^0 = \frac{3!}{1 \cdot 3!} x^3 \cdot 1 = 1 \cdot x^3 = x^3$$

For $k=1$:

$$\binom{3}{1} x^{3-1} y^1 = \frac{3!}{1!(3-1)!} x^2 y^1 = \frac{3!}{1!2!} x^2 y = \frac{3 \times 2 \times 1}{(1)(2 \times 1)} x^2 y = 3 x^2 y$$

For $k=2$:

$$\binom{3}{2} x^{3-2} y^2 = \frac{3!}{2!(3-2)!} x^1 y^2 = \frac{3!}{2!1!} x y^2 = \frac{3 \times 2 \times 1}{(2 \times 1)(1)} x y^2 = 3 x y^2$$

For $k=3$:

$$\binom{3}{3} x^{3-3} y^3 = \frac{3!}{3!(3-3)!} x^0 y^3 = \frac{3!}{3!0!} x^0 y^3 = \frac{3!}{3! \cdot 1} \cdot 1 \cdot y^3 = 1 \cdot y^3 = y^3$$

Adding these terms together, we get the full expansion:

$$(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$$

Some key properties of binomial expansions:

• The expansion of $(a+b)^n$ has $n+1$ terms.
• The sum of the exponents of $a$ and $b$ in each term is always $n$.
• The binomial coefficients are symmetric; for example, $\binom{n}{k} = \binom{n}{n-k}$.