Binomial Theorem and Pascal's Triangle

By the end of this lesson, you will be able to expand binomials using the Binomial Theorem, apply Pascal's Triangle, find specific terms, and prove binomial coefficient identities using induction.

Review: The Binomial Theorem

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

$\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$ — "n choose k"

Pascal's Triangle and the Identity

Pascal's Identity: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$

This is the rule for building Pascal's Triangle: each entry is the sum of the two above it.

Proof using induction: This can be proved directly from the combinatorial definition.

Binomial Coefficient Identities

Symmetry: $\binom{n}{k} = \binom{n}{n-k}$

Row sum: $\sum_{k=0}^{n}\binom{n}{k} = 2^n$ (set $a = b = 1$ in the theorem)

Alternating sum: $\sum_{k=0}^{n}(-1)^k\binom{n}{k} = 0$ (set $a = 1, b = -1$)

Example 1 — Row Sum Proof

Prove $\sum_{k=0}^{n}\binom{n}{k} = 2^n$.

$(1+1)^n = \sum_{k=0}^{n}\binom{n}{k}1^{n-k}1^k = \sum_{k=0}^{n}\binom{n}{k} = 2^n$ ✓

Example 2 — Proving a Coefficient Identity by Induction

Prove Pascal's Identity: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$

Direct proof (algebraic):

$\binom{n-1}{k-1} + \binom{n-1}{k} = \frac{(n-1)!}{(k-1)!(n-k)!} + \frac{(n-1)!}{k!(n-k-1)!}$

$= \frac{(n-1)!}{(k-1)!(n-k-1)!}\left[\frac{1}{n-k} + \frac{1}{k}\right] = \frac{(n-1)!}{(k-1)!(n-k-1)!}\cdot\frac{n}{k(n-k)} = \frac{n!}{k!(n-k)!} = \binom{n}{k}$ ✓

Multinomial Theorem (Extension)

$(a + b + c)^n$ has terms of the form $\dfrac{n!}{i!j!k!}a^i b^j c^k$ where $i+j+k = n$.

Example: $(a+b+c)^2 = a^2+b^2+c^2+2ab+2ac+2bc$

Coefficients: $\dfrac{2!}{2!0!0!}=1$, $\dfrac{2!}{1!1!0!}=2$, etc.

Finding the Term with a Specific Power

In $(3x - 2y)^8$, find the term containing $x^5y^3$.

$k = 3$ (power of $(-2y)$ is $k = 3$):

$\binom{8}{3}(3x)^5(-2y)^3 = 56 \cdot 243x^5 \cdot (-8y^3) = -108864x^5y^3$

Summary

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