5.4 The Binomial Theorem & Expansions

5.4 The Binomial Theorem & Expansions

Expanding powers of a binomial expression $(a + b)^n$ by repeated polynomial multiplication quickly becomes computationally inefficient for higher exponents $n$. The Binomial Theorem solves this challenge by providing a direct algebraic formula for expanding any power of a binomial using binomial coefficients and combinatorial identities.

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Combinations & Binomial Coefficients

The binomial coefficient $\binom{n}{r}$ (read "$n$ choose $r$") represents the number of ways to select a subset of $r$ un-ordered elements from a set of $n$ distinct elements.

$\binom{n}{r} = {}_n C_r = \frac{n!}{r!(n - r)!} \quad \text{for } 0 \le r \le n$

Fundamental Algebraic Properties

1. Boundary Values: $\binom{n}{0} = \frac{n!}{0! n!} = 1$ and $\binom{n}{n} = \frac{n!}{n! 0!} = 1$.
2. Symmetry Identity: $\binom{n}{r} = \binom{n}{n - r}$.
3. Pascal's Identity: $\binom{n}{r} + \binom{n}{r - 1} = \binom{n + 1}{r}$.

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Pascal's Triangle

Pascal's Triangle is a triangular array of numbers where row $n$ lists the binomial coefficients $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$. Each interior number is formed by adding the two numbers directly above it (reflecting Pascal's Identity).

Row 0:                1
Row 1:              1   1
Row 2:            1   2   1
Row 3:          1   3   3   1
Row 4:        1   4   6   4   1
Row 5:      1   5  10  10   5   1
Row 6:    1   6  15  20  15   6   1

Key Structural Properties

• Row Sum Identity: The sum of all binomial coefficients in row $n$ equals $2^n$:

$\sum_{k=0}^n \binom{n}{k} = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n$

• Symmetric Balance: Each row reads identically forward and backward.

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The Binomial Theorem Formula

For any real numbers $a, b$ and non-negative integer $n$:

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

Key Expansion Rules

1. Total Number of Terms: Expanding $(a + b)^n$ produces exactly $n + 1$ terms.
2. Exponent Behavior: The exponent of $a$ starts at $n$ and decreases by $1$ in each term down to $0$. The exponent of $b$ starts at $0$ and increases by $1$ in each term up to $n$.
3. Homogeneity: The sum of exponents of $a$ and $b$ in every term is constant and equals $n$ ($n - k + k = n$).
4. Alternating Signs: When expanding $(a - b)^n$, replace $b$ with $(-b)$. Terms with odd powers of $b$ become negative, causing signs to alternate: $+ , - , + , - \dots$

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Finding a Specific Term (General Term Formula)

Instead of computing the complete polynomial expansion, any individual term in $(a + b)^n$ can be targeted using the general term formula for the $(k + 1)$-th term:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

Important Indexing Rule: The index $k$ is always $1$ less than the term position number. For instance, to find the 4th term ($T_4$), substitute $k = 3$.

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Comprehensive Worked Examples

Worked Example 1: Calculating Binomial Coefficients

Problem: Calculate the binomial coefficient $\binom{8}{3}$, state its symmetric counterpart, and verify row symmetry.

Step-by-Step Solution:

1. Apply combination formula: $\binom{8}{3} = \frac{8!}{3!(8 - 3)!} = \frac{8!}{3! 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$
2. Apply symmetry property $\binom{n}{r} = \binom{n}{n - r}$: $\binom{8}{3} = \binom{8}{8 - 3} = \binom{8}{5} = 56$

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Worked Example 2: Step-by-Step Expansion of $(2x - y)^5$

Problem: Expand $(2x - y)^5$ fully using the Binomial Theorem.

Step-by-Step Solution:

1. Retrieve row 5 coefficients from Pascal's Triangle: $1, 5, 10, 10, 5, 1$.
2. Set up general terms with $a = 2x$ and $b = -y$:
   • $k=0: \binom{5}{0} (2x)^5 (-y)^0 = 1 \cdot (32x^5) \cdot 1 = 32x^5$
   • $k=1: \binom{5}{1} (2x)^4 (-y)^1 = 5 \cdot (16x^4) \cdot (-y) = -80x^4 y$
   • $k=2: \binom{5}{2} (2x)^3 (-y)^2 = 10 \cdot (8x^3) \cdot (y^2) = 80x^3 y^2$
   • $k=3: \binom{5}{3} (2x)^2 (-y)^3 = 10 \cdot (4x^2) \cdot (-y^3) = -40x^2 y^3$
   • $k=4: \binom{5}{4} (2x)^1 (-y)^4 = 5 \cdot (2x) \cdot (y^4) = 10x y^4$
   • $k=5: \binom{5}{5} (2x)^0 (-y)^5 = 1 \cdot 1 \cdot (-y^5) = -y^5$
3. Combine terms into expanded polynomial: $(2x - y)^5 = 32x^5 - 80x^4 y + 80x^3 y^2 - 40x^2 y^3 + 10xy^4 - y^5$

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Worked Example 3: Finding a Constant Term

Problem: Find the constant term (the term independent of $x$) in the expansion of $\left( 2x^3 - \frac{1}{x} \right)^4$.

Step-by-Step Solution:

1. Write general term $T_{k+1}$ with $a = 2x^3$, $b = -x^{-1}$, and $n = 4$: $T_{k+1} = \binom{4}{k} \left( 2x^3 \right)^{4-k} \left( -x^{-1} \right)^k$
2. Separate scalar numbers from powers of $x$: $T_{k+1} = \binom{4}{k} 2^{4-k} (-1)^k \cdot x^{3(4-k)} x^{-k} = \binom{4}{k} 2^{4-k} (-1)^k \cdot x^{12 - 3k - k} = \binom{4}{k} 2^{4-k} (-1)^k \cdot x^{12 - 4k}$
3. Set the exponent of $x$ equal to zero for the constant term: $12 - 4k = 0 \implies 4k = 12 \implies k = 3$
4. Substitute $k = 3$ into scalar terms to calculate constant value: $T_4 = \binom{4}{3} 2^{4-3} (-1)^3 = 4 \cdot 2^1 \cdot (-1) = 4 \cdot 2 \cdot (-1) = -8$ Conclusion: The constant term is $-8$.