Binomial Theorem

Expanding powers of a two-term algebraic expression (binomial) is a fundamental operation in algebra. The geometric arrangement of binomial coefficients reveals profound recursive structures.

Centuries before Blaise Pascal published his Traité du triangle arithmétique in 1654, the ancient Indian mathematician Pingala described the exact triangular array of combinatoric numbers in his Sanskrit treatise Chandas Shastra (c. 3rd–2nd century BCE) under the name Meru Prastara (the staircase/pyramid of Mount Meru). Later, Persian mathematician Omar Khayyam (c. 1048–1131) and Chinese mathematician Yang Hui (1261) documented the same pyramidal arithmetic array.

Each row begins and ends with $1$. Every interior entry is the exact arithmetic sum of the two numbers directly above it in the preceding row:

Algebraic Proof:

Express factorials with common denominators: $r! = r \cdot (r - 1)!$ and $(n - r + 1)! = (n - r + 1)(n - r)!$:

The Binomial Theorem provides an exact algebraic formula to expand $(a + b)^n$ into a sum of monomials without expanding brackets term-by-term.

Let $P(n)$ be the statement: $(a + b)^n = \sum_{r=0}^n {^nC_r} a^{n-r} b^r$.

1. Base Step ($n = 1$): $$\text{LHS} = (a + b)^1 = a + b$$ $$\text{RHS} = {^1C_0} a^1 b^0 + {^1C_1} a^0 b^1 = 1 \cdot a + 1 \cdot b = a + b$$ Since $\text{LHS} = \text{RHS}$, $P(1)$ is strictly true.
2. Inductive Hypothesis: Assume $P(k)$ is true for some positive integer $k \ge 1$: $$(a + b)^k = {^kC_0} a^k + {^kC_1} a^{k-1} b + {^kC_2} a^{k-2} b^2 + \dots + {^kC_k} b^k$$
3. Inductive Step ($n = k + 1$): Multiply both sides by $(a + b)$: $$(a + b)^{k+1} = (a + b) (a + b)^k = (a + b) \left[ {^kC_0} a^k + {^kC_1} a^{k-1} b + \dots + {^kC_k} b^k \right]$$ Distribute $a$ and $b$: $$= a \left[ \sum_{r=0}^k {^kC_r} a^{k-r} b^r \right] + b \left[ \sum_{r=0}^k {^kC_r} a^{k-r} b^r \right]$$ $$= \sum_{r=0}^k {^kC_r} a^{k-r+1} b^r + \sum_{r=0}^k {^kC_r} a^{k-r} b^{r+1}$$ Expand and group like powers of $a^{k+1-r} b^r$: $$= {^kC_0} a^{k+1} + \sum_{r=1}^k \left( {^kC_r} + {^kC_{r-1}} \right) a^{k+1-r} b^r + {^kC_k} b^{k+1}$$ Using Pascal\'s Identity (${^kC_r} + {^kC_{r-1}} = {^{k+1}C_r}$) and noting ${^kC_0} = 1 = {^{k+1}C_0}$ and ${^kC_k} = 1 = {^{k+1}C_{k+1}}$: $$= {^{k+1}C_0} a^{k+1} + \sum_{r=1}^k {^{k+1}C_r} a^{k+1-r} b^r + {^{k+1}C_{k+1}} b^{k+1} = \sum_{r=0}^{k+1} {^{k+1}C_r} a^{k+1-r} b^r$$ This proves $P(k + 1)$ is true whenever $P(k)$ is true.

Hence, by the Principle of Mathematical Induction, the theorem holds for all $n \in \mathbb{N}$.

• Total Terms: The expansion contains exactly $\mathbf{n + 1}$ terms. For instance, $(x + y)^{10}$ has 11 terms.
• Degree Invariance: In every term ${^nC_r} a^{n-r} b^r$, the sum of the exponents of $a$ and $b$ is $(n - r) + r = \mathbf{n}$.
• Symmetry of Coefficients: Coefficients equidistant from the beginning and end are equal: ${^nC_r} = {^nC_{n-r}}$.
• Alternating Expansion $(a - b)^n$: $$(a - b)^n = [a + (-b)]^n = \sum_{r=0}^n (-1)^r {^nC_r} a^{n-r} b^r = {^nC_0} a^n - {^nC_1} a^{n-1} b + {^nC_2} a^{n-2} b^2 - \dots + (-1)^n {^nC_n} b^n$$
• Standard Single-Variable Expansion $(1 + x)^n$: $$(1 + x)^n = {^nC_0} + {^nC_1} x + {^nC_2} x^2 + \dots + {^nC_n} x^n = \sum_{r=0}^n {^nC_r} x^r$$

To analyze specific terms in an expansion without writing out the full polynomial, mathematicians utilize the General Term formula.

Because the first term corresponds to $r = 0$, the $(r + 1)$-th term in the expansion of $(a + b)^n$ is denoted $T_{r+1}$:

For the alternating expansion $(a - b)^n$:

In the expansion of $(a + b)^n$, which contains $(n + 1)$ terms:

Alternative Method: The $p$-th term from the end of $(a + b)^n$ is simply the $p$-th term from the beginning of the inverted expansion $(b + a)^n$.

The number of middle terms depends strictly on whether the exponent $n$ is even or odd:

Two classic high-weightage problem types in WBCHSE and competitive examinations are finding the constant term and the numerically greatest term.

To find the term independent of $x$ in an expansion such as $\left(A x^p + \frac{B}{x^q}\right)^n$:

1. Write the general term $T_{r+1}$: $$T_{r+1} = {^nC_r} (A x^p)^{n-r} \left(\frac{B}{x^q}\right)^r = {^nC_r} A^{n-r} B^r \cdot x^{p(n-r) - qr} = {^nC_r} A^{n-r} B^r \cdot x^{pn - (p+q)r}$$
2. Set the net exponent of $x$ equal to zero: $$pn - (p + q)r = 0 \implies \mathbf{r = \frac{pn}{p + q}}$$
3. Integrity Check: If $r$ evaluates to a non-negative integer with $0 \le r \le n$, substitute $r$ back into $T_{r+1}$ to obtain the constant term. If $r$ is fractional or negative, no term independent of $x$ exists in the expansion.

To determine which term has the largest numerical value in the expansion of $(1 + x)^n$ for a specified real value of $x$:

• Case 1: If $\frac{(n + 1)|x|}{|x| + 1} = m$ is an integer: Both the $m$-th term ($T_m$) and $(m + 1)$-th term ($T_{m+1}$) are numerically equal and are the two greatest terms.
• Case 2: If $\frac{(n + 1)|x|}{|x| + 1} = m + f$ is not an integer: (where $m = \lfloor \cdot \rfloor$ is the integer part), then the single numerically greatest term is $\mathbf{T_{m+1}}$.

For brevity, the binomial coefficient ${^nC_r}$ is often written simply as $C_r$. These coefficients satisfy fundamental summation identities.

To evaluate sums where coefficients are multiplied by integers, two primary methods are available:

• Identity 1: $r \cdot {^nC_r} = n \cdot {^{n-1}C_{r-1}}$: $$\sum_{r=1}^n r C_r = \sum_{r=1}^n n \cdot {^{n-1}C_{r-1}} = n \sum_{r=1}^n {^{n-1}C_{r-1}} = \mathbf{n \cdot 2^{n-1}}$$ Alternative Calculus Method: Differentiate $(1 + x)^n = \sum_{r=0}^n C_r x^r$ with respect to $x$: $$n(1 + x)^{n-1} = \sum_{r=1}^n r C_r x^{r-1} \xrightarrow{x = 1} \mathbf{\sum_{r=1}^n r C_r = n \cdot 2^{n-1}}$$
• Identity 2: Sum of Squares of Coefficients: $$C_0^2 + C_1^2 + C_2^2 + \dots + C_n^2 = \mathbf{^{2n}C_n = \frac{(2n)!}{(n!)^2}}$$ Proof: Equate the coefficient of $x^n$ on both sides of $(1 + x)^n (x + 1)^n = (1 + x)^{2n}$, noting that $C_r = C_{n-r}$.
• Identity 3: Integration of Binomial Series: $$\int_0^1 (1 + x)^n dx = \sum_{r=0}^n \frac{C_r}{r + 1} \implies \left[ \frac{(1 + x)^{n+1}}{n + 1} \right]_0^1 = \mathbf{\frac{2^{n+1} - 1}{n + 1} = \sum_{r=0}^n \frac{C_r}{r + 1}}$$

When the exponent is not a positive integer, the expansion does not terminate; it becomes an infinite converging series with strict domain constraints.

1. Infinite Terms: The series never terminates because $n(n - 1)\dots(n - r + 1)$ never equals zero for non-integer $n$.
2. Convergence Condition: The expansion is strictly valid if and only if $|x| < 1$. If $|x| \ge 1$, the series diverges and is invalid.
3. Normalization to Unity: To expand $(a + b)^n$ where $n \notin \mathbb{N}$, one must factor out $a^n$: $$(a + b)^n = a^n \left(1 + \frac{b}{a}\right)^n, \quad \text{valid only when } \left|\frac{b}{a}\right| < 1 \iff |b| < |a|$$

When $|x|$ is very small ($x^2, x^3 \approx 0$):

Example: To compute $\sqrt{1.02} = (1 + 0.02)^{1/2}$: