📋 Prerequisites

• Permutations & Combinations
• Polynomials & Factoring

The binomial theorem tells you how to expand (a + b)\(^{n}\) without multiplying it all out. The coefficients are binomial coefficients from Pascal's triangle. Beyond just expansion, the theorem gives you powerful identities and techniques for solving competition problems.

📚 Key Concepts

The Binomial Theorem

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

Expanded form:

\[ (a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + ... + \binom{n}{n}b^n \]

The coefficient of \(a^{n-k}\(\)b^{k} \(is \binom{n}{k},\) the k-th entry in row n of Pascal's triangle.\)

Why It Works (Combinatorial Proof)

(a + b)\(^{n}\) = (a+b)(a+b)...(a+b) [n factors]

To get a term with \(b^{k},\) you pick b from k of the n factors and a from the rest.

\(Number of ways: \binom{n}{k}. So coefficient = \binom{n}{k}.\)

This is the most satisfying proof — it tells you WHY the coefficients are binomial coefficients.

Pascal's Triangle Revisited

Row n of Pascal's triangle = coefficients of (a+b)\(^{n}.\)

n=0:       1
n=1:      1 1
n=2:     1 2 1
n=3:    1 3 3 1
n=4:   1 4 6 4 1
n=5:  1 5 10 10 5 1
      

\(Pascal's identity \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} corresponds to how each row builds from the one above.\)

Finding a Specific Term

Term with \(b^{k}\) (or the (k+1)-th term):

\[ \binom{n}{k} a^{n-k} b^k \]

\(Example: coefficient of x^{3} in (2+x)^{5} = \binom{5}{3} \\times 2^{2} \\times 1^{3} = 10 \\times 4 = 40\)

Common trick: for (ax + by)\(^{n},\) don't forget the coefficients a and b each get raised to their powers too!

Substitution Tricks

Plug in specific values for a and b to get identities.

Set a = 1, b = 1: \((2^{n} = \\sum \binom{n}{k}\)\) — sum of row n = \(2^{n}\) (total subsets)

Set a = 1, b = -1: \(0 = \\sum (-1)^{k}\binom{n}{k}\) — alternating sum = 0 (for n > 0)

Set a = 1, b = 2: \((3^{n} = \\sum 2^{k}\)\binom{n}{k}\)

This technique generates all kinds of combinatorial identities.

Vandermonde's Identity

\[ \binom{m+n}{k} = \sum_{i=0}^{k} \binom{m}{i} \binom{n}{k-i} \]

Combinatorial proof: choose k items from m+n total. Split into group of m and group of n. Pick i from first group, k-i from second. Sum over all i.

Algebraic proof: coefficient of \(x^{k} in (1+x)^{m+n} = (1+x)^{m}(1+x)^{n}.\)

Hockey-Stick Identity

\[ \sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1} \]

Name comes from the shape on Pascal's triangle: a diagonal of numbers sums to the number just below the end of the diagonal.

Very useful for summing binomial coefficients along a diagonal.

Multinomial Theorem (Generalization)

\(For more than two terms: \((a_{1} + a_{2}\) + ... + a_{k})\(^{n}\)\)

\(\(Coefficient of \(a_{1}\)(^{n_{1}}a_{2}\)(^{n_{2}}...a\)_{k}\(^{n_{k}}\) where \(n_{1}+...+n\)_{k} = \)n:\)

\[ \binom{n}{n_1, n_2, ..., n_k} = \frac{n!}{n_1! n_2! ... n_k!} \]

This is the same as the permutations-with-repetition formula — makes sense combinatorially.

Applications

• Expanding polynomials quickly
• Finding specific coefficients without full expansion
• Proving combinatorial identities
• Binomial probability distribution
• \(Approximations: (1 + x)^{n} \\approx 1 + nx for small x\)

✏️ Example Problems