18.2 Binomial Theorem, Pascal's Triangle & General Term Determination

18.2 Binomial Theorem, Pascal's Triangle & General Term Determination

Combinatorial Foundations: Combinations & Factorial Notation

The algebraic expansion of polynomial powers is intimately connected to combinatorics. The number of distinct subsets of size $k$ that can be selected from an $n$-element set without regard to order is given by the combination formula, denoted $\binom{n}{k}$ (read '$n$ choose $k$') or ${}_nC_k$:

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

where $n! = n \times (n - 1) \times \dots \times 2 \times 1$, with the fundamental convention $0! = 1$. The combination values satisfy two structural algebraic properties:

1. Combinatorial Symmetry: Selecting $k$ items to include is equivalent to selecting $n - k$ items to exclude: $$\binom{n}{k} = \binom{n}{n - k}$$
2. Boundary Values: For any non-negative integer $n$, $\binom{n}{0} = 1$ and $\binom{n}{n} = 1$.

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Pascal's Triangle & Pascal's Identity

When combinations are arranged sequentially by row $n$ and entry $k$, they form Pascal's Triangle. The generating mechanism that produces each interior entry by adding the two adjacent numbers directly above it is formalized as Pascal's Identity:

$$\binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k} \quad \text{for } 1 \le k \le n - 1$$

Algebraic proof of Pascal's Identity verifies this relationship:

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

Key geometric and algebraic invariants of Pascal's Triangle include:

• Row Sums: The sum of entries in row $n$ equals $2^n$: $\sum_{k=0}^n \binom{n}{k} = 2^n$. This represents the total number of subsets of an $n$-element set (the cardinality of the power set).
• Alternating Row Sums: For any row $n \ge 1$, the alternating sum of entries is identically zero: $\sum_{k=0}^n (-1)^k \binom{n}{k} = 0$.

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The Binomial Theorem: Formulation & Properties

For any real or complex numbers $a$ and $b$ and any non-negative integer $n$, the Binomial Theorem provides the closed-form expansion of $(a + b)^n$:

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

Expansion properties tested on the FTCE examination include:

• Term Count: The complete expansion contains exactly $n + 1$ terms.
• Homogeneous Total Degree: The sum of exponents of $a$ and $b$ in every individual term is constant: $(n - k) + k = n$.
• Sum of All Coefficients: Setting all variable terms to 1 evaluates the sum of coefficients instantaneously: substituting $x = 1, y = 1$ into $(c_1 x + c_2 y)^n$ yields $(c_1 + c_2)^n$.

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The General $(k+1)$-th Term Formula

Because the summation index starts at $k = 0$, the terms are indexed sequentially:

• $1$st term ($k = 0$): $T_1 = \binom{n}{0} a^n b^0$
• $2$nd term ($k = 1$): $T_2 = \binom{n}{1} a^{n-1} b^1$
• In general, the $(k+1)$-th term is given by: $$T_{k+1} = \binom{n}{k} a^{n-k} b^k$$

A ubiquitous candidate error is confusing the term order $m$ with the index parameter $k$. To find the 6th term, one must set $k = 6 - 1 = 5$, not $k = 6$.

When expressions involve negative signs, fractions, or algebraic powers, each component must be partitioned carefully into its numerical coefficient and variable exponent:

$$\left(A x^p + B x^{-q}\right)^n \implies T_{k+1} = \binom{n}{k} (A x^p)^{n-k} (B x^{-q})^k = \binom{n}{k} A^{n-k} B^k x^{p(n-k) - q k}$$

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Finding Specific Terms & Constant Terms

To identify a term possessing a specific power of $x$ (such as $x^m$, or the constant term where $x^0 = 1$):

1. Write out the general term $T_{k+1}$ with explicit powers.
2. Collect the variable exponents into a single linear equation in $k$: $p(n - k) - q k = m$.
3. Solve for $k$. If $k$ is an integer in the domain $0 \le k \le n$, the term exists; if $k$ is fractional or outside the domain, the coefficient of that power is 0.
4. Substitute the solved integer $k$ back into the coefficient expression $\binom{n}{k} A^{n-k} B^k$ to evaluate the exact numerical value.

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Combinatorics & the Binomial Probability Distribution

The Binomial Theorem provides the mathematical foundation for the binomial probability distribution. In an experiment consisting of $n$ independent and identical Bernoulli trials, where each trial results in either 'success' with probability $p$ or 'failure' with probability $q = 1 - p$, the probability of obtaining exactly $k$ successes is given by:

$$P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} \quad \text{for } k \in \{0, 1, 2, \dots, n\}$$

The total probability across all possible outcomes sums to 1 as an immediate consequence of the Binomial Theorem:

$$\sum_{k=0}^n P(X = k) = \sum_{k=0}^n \binom{n}{k} p^k (1 - p)^{n-k} = (p + (1 - p))^n = 1^n = 1$$

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Pascal's Triangle & Binomial Coefficients Table

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Worked Exemplar: Determining Specific Terms in a Rational Expansion

Problem: Consider the binomial expansion of $\left(2x^2 - \frac{3}{x}\right)^9$.

1. Determine the term independent of $x$ (the constant term).
2. Determine the coefficient of $x^6$.

Step 1: Set Up the General Term Formula. Identify $a = 2x^2$, $b = -3x^{-1}$, and $n = 9$. The general $(k+1)$-th term is:

$$T_{k+1} = \binom{9}{k} (2x^2)^{9-k} \left(-\frac{3}{x}\right)^k = \binom{9}{k} 2^{9-k} (x^2)^{9-k} (-3)^k (x^{-1})^k$$

Separate the numerical and variable factors:

$$T_{k+1} = \binom{9}{k} 2^{9-k} (-3)^k x^{2(9-k) - k} = \binom{9}{k} 2^{9-k} (-3)^k x^{18 - 3k}$$

Step 2: Solve for the Constant Term ($x^0$). The constant term corresponds to an exponent of zero on $x$:

$$18 - 3k = 0 \implies 3k = 18 \implies k = 6$$

Because $k = 6$ is an integer satisfying $0 \le 6 \le 9$, the constant term exists and corresponds to the $(6 + 1) = 7\text{th}$ term ($T_7$). Evaluate the coefficient:

$$\text{Constant Term} = \binom{9}{6} 2^{9-6} (-3)^6 = \binom{9}{3} 2^3 (-3)^6$$

Compute each component:

$$\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$ $$2^3 = 8, \quad (-3)^6 = 729$$ $$\text{Constant Term} = 84 \times 8 \times 729 = 672 \times 729 = 489,888$$

Step 3: Solve for the Coefficient of $x^6$. Set the variable exponent equal to 6:

$$18 - 3k = 6 \implies 3k = 12 \implies k = 4$$

This corresponds to the 5th term ($T_5$). Evaluate the coefficient:

$$\text{Coefficient of } x^6 = \binom{9}{4} 2^{9-4} (-3)^4 = \binom{9}{4} 2^5 (-3)^4$$

Compute each component:

$$\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$$ $$2^5 = 32, \quad (-3)^4 = 81$$ $$\text{Coefficient of } x^6 = 126 \times 32 \times 81 = 4,032 \times 81 = 326,592$$

Both specific terms are completely determined.