Binomial Theorem

Binomial Expansion for Positive Integer Index

The Binomial Theorem expands $(a+x)^n$ into a sum of $n+1$ terms, where each term's coefficient is a binomial coefficient from Pascal's Triangle.

$$(a+x)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} x^r$$

Each term combines a binomial coefficient with decreasing powers of $a$ and increasing powers of $x$.

$\binom{n}{r}$=Binomial coefficient — number of ways to choose $r$ items from $n$

$a^{n-r}$=First term raised to a decreasing power

$x^r$=Second term raised to an increasing power

$a = 1$

→

$(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n$

$x$ replaced by $-x$

→

Signs alternate: $(a-x)^n = \binom{n}{0}a^n - \binom{n}{1}a^{n-1}x + \binom{n}{2}a^{n-2}x^2 - \dots$

Term Count: The expansion of $(a+x)^n$ always has exactly $n+1$ terms.

Exponent Sum Rule: In every term, the exponents of $a$ and $x$ always add up to $n$.

Coefficient Symmetry: $\binom{n}{r} = \binom{n}{n-r}$, so the first and last coefficients match, the second and second-to-last match, etc.

Pascal's Identity: $\binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r}$ — this is how each entry in Pascal's Triangle is formed from the row above.

The General Term ($T_{r+1}$)

The General Term formula gives any specific term in an expansion directly, without expanding the entire binomial.

$$T_{r+1} = \binom{n}{r} a^{n-r} x^r$$

The $(r+1)$-th term of $(a+x)^n$, counting from the beginning.

$T_{r+1}$=The term at position $r+1$ in the expansion

$\binom{n}{r}$=Binomial coefficient $\frac{n!}{r!(n-r)!}$

Finding the $k$-th term from the end

→

It equals the $(n - k + 2)$-th term from the beginning

Index Rule: The $r$ value is always one less than the term number. For the 5th term, use $r = 4$.

Term from End: The $k$-th term from the end in $(a+x)^n$ is the $(n-k+2)$-th term from the beginning.

Coefficient vs Term: The 'coefficient of $x^p$' means only the numerical factor — strip out all powers of $x$ after substituting $r$.

Finding Specific Terms

A term is 'independent of $x$' (or a constant term) when the combined power of $x$ in the General Term equals zero.

Step 1: Write $T_{r+1}$ for the binomial, keeping all $x$-terms explicit.

Step 2: Consolidate all $x$ factors into a single power $x^{f(r)}$ where $f(r)$ is an expression in $r$.

Step 3: Set $f(r) = 0$ and solve for $r$. If $r$ is not a non-negative integer $\leq n$, no independent term exists.

Step 4: Substitute the integer $r$ back to compute the coefficient.

Finding term involving $x^p$: Same process but set $f(r) = p$ instead of $0$.

The number of middle terms in an expansion depends on whether $n$ is even or odd.

Even $n$: Exactly one middle term at position $\left(\frac{n}{2} + 1\right)$. For $(a+x)^{12}$, the middle term is $T_7$.

Odd $n$: Two middle terms at positions $\frac{n+1}{2}$ and $\frac{n+3}{2}$. For $(a+x)^{11}$, middle terms are $T_6$ and $T_7$.

Quick Check: Total terms $= n+1$. If odd (even $n$), one middle; if even (odd $n$), two middles.

Coefficient Sum Properties

Substituting specific values of $a$ and $x$ into the expansion produces powerful identities about binomial coefficients.

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

Setting $a = 1, x = 1$ in $(a+x)^n$ gives the sum of all binomial coefficients.

$2^n$=Total sum of all binomial coefficients for index $n$

$a = 1, x = -1$

→

Alternating sum $= 0$: $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots = 0$

Sum of All Coefficients: Put $a=1, x=1$: sum $= 2^n$.

Alternating Sum: Put $a=1, x=-1$: alternating sum $= 0$.

Even-Position = Odd-Position: $\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \dots = \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots = 2^{n-1}$.

Weighted Sum: $\binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \dots + n\binom{n}{n} = n \cdot 2^{n-1}$.

Binomial Series (Fractional and Negative Index)

When the index $n$ is a negative integer or a fraction, the Binomial Series gives an infinite expansion valid only when $|x| < 1$.

$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$$

An infinite series that converges when $|x| < 1$. Unlike the positive integer case, it does not terminate.

$n$=The index — can be negative or fractional (e.g., $-1$, $\frac{1}{2}$, $-\frac{2}{3}$)

$|x| < 1$=Convergence condition — the series is only valid when the absolute value of $x$ is less than 1

$n = -1$

→

$(1+x)^{-1} = 1 - x + x^2 - x^3 + \dots$

$n = -2$

→

$(1+x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \dots$

$n = \frac{1}{2}$

→

$(1+x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \dots$

Key Difference: For positive integer $n$, the expansion is finite ($n+1$ terms). For fractional/negative $n$, it is infinite.

$\binom{n}{r}$ Not Valid: The standard $\binom{n}{r}$ notation is meaningless for non-positive-integer $n$. Use the product form $\frac{n(n-1)(n-2)\dots(n-r+1)}{r!}$ instead.

Convergence Condition: The base must be written as $(1 + \text{something})^n$ where $|\text{something}| < 1$.

Factoring Technique: To expand $(a + bx)^n$ where $a \neq 1$, factor out $a^n$: $(a+bx)^n = a^n\left(1 + \frac{bx}{a}\right)^n$, then apply the series to $\left(1 + \frac{bx}{a}\right)^n$.

$(1-x)^{-1}$ Series: $(1-x)^{-1} = 1 + x + x^2 + x^3 + \dots$ — this is the geometric series.

The general term of the binomial series uses descending products instead of factorials.

$$T_{r+1} = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!} x^r$$

Negative Index Pattern: For $(1+x)^{-m}$, the general term is $(-1)^r \binom{m+r-1}{r} x^r$.

Sign Pattern: When $n < 0$, the factors in the numerator alternate sign, producing an alternating series.

Approximation: Since $|x| < 1$ and higher powers shrink, often the first 3–4 terms give sufficient accuracy to 3 decimal places.

Applications: Approximation and Series Summation

The binomial series is used to approximate roots and powers of numbers close to perfect powers.

Strategy: Express the number as $(\text{perfect power})(1 \pm \text{small quantity})^n$, then expand and keep first few terms.

Example Pattern: $\sqrt[3]{30} = (27+3)^{1/3} = 3(1+\frac{1}{9})^{1/3} \approx 3\left(1 + \frac{1}{27} - \frac{1}{729}\right) \approx 3.107$.

Neglecting Higher Powers: When $x$ is very small, $(1+x)^n \approx 1 + nx$ (first-order approximation).

Second-Order: $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2$ when cube and higher powers are negligible.

Infinite series can be identified as binomial expansions by comparing terms with the standard binomial series pattern.

Method: Compare the given series with $1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$ to find $n$ and $x$.

Step 1: Set $nx = $ (second term of series) to get a relation between $n$ and $x$.

Step 2: Set $\frac{n(n-1)}{2!}x^2 = $ (third term) and solve simultaneously for $n$ and $x$.

Step 3: The sum of the series is $(1+x)^n$ with the found values of $n$ and $x$.