The binomial theorem and Pascal's triangle: expansion of $(a + b)^n$ and the general term

What this dot point is asking

NESA wants you to expand $(a + b)^n$ using the binomial theorem, identify the general term and specific powered terms, find coefficients (including the independent or constant term), and use Pascal's triangle for small $n$. The two halves of the dot point are tightly linked: the entries of Pascal's triangle are the binomial coefficients $\binom{n}{k}$, so understanding one explains the other. The workhorse for almost every exam question is the general term $T_{k+1}$.

The answer

The binomial theorem

For any non-negative integer $n$,

Each term has a coefficient $\binom{n}{k}$ and powers of $a$ and $b$ that always add to $n$. The reason the coefficient is $\binom{n}{k}$ is itself a counting argument: expanding $(a+b)^n = (a+b)(a+b)\cdots(a+b)$ means choosing $a$ or $b$ from each of the $n$ brackets; a term $a^{n-k}b^k$ comes from choosing $b$ in exactly $k$ of the brackets, and there are $\binom{n}{k}$ ways to do that. So the binomial theorem is combinations in disguise, which is why this dot point sits in the combinatorics module.

Expanded form for small $n$

The coefficients $1, 2, 1$; $1, 3, 3, 1$; $1, 4, 6, 4, 1$; $1, 5, 10, 10, 5, 1$ are exactly the rows of Pascal's triangle, and the powers of $a$ count down from $n$ to $0$ while the powers of $b$ count up from $0$ to $n$.

The general term

The $(k + 1)$-th term in the expansion of $(a + b)^n$ is

So $T_1 = a^n$ (with $k = 0$), $T_2 = n a^{n - 1} b$, and the last term is $T_{n + 1} = b^n$. The general term is the single most useful object in this topic: "find the coefficient of $x^m$" and "find the term independent of $x$" both reduce to writing $T_{k+1}$, finding which $k$ produces the power you want, and evaluating.

Pascal's triangle, built row by row

Each row of Pascal's triangle gives the coefficients of $(a + b)^n$ for that $n$. The triangle is built from two simple rules: every row begins and ends with $1$, and every interior entry is the sum of the two entries directly above it. That second rule is Pascal's rule, $\binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k}$, so the triangle is a fast way to generate binomial coefficients for small $n$ without any factorials.

Stage 1, the apex and the edges. Start with a single $1$ at the top (this is $\binom{0}{0}$, the coefficient in $(a+b)^0 = 1$). Each new row begins and ends with $1$, because $\binom{n}{0} = \binom{n}{n} = 1$.

Stage 2, add the two above. To make the next row, each interior entry is the sum of the two entries diagonally above it. In row $3$ the middle entry is $1 + 2 = 3$. This is Pascal's rule in action.

Stage 3, the triangle to row 5. Continuing the same rule fills out every row. Row $5$ reads $1, 5, 10, 10, 5, 1$, and these are precisely the coefficients in $(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$.

Reading the rows ($n$ starting from $0$):

• $n = 0$: $1$
• $n = 1$: $1, 1$
• $n = 2$: $1, 2, 1$
• $n = 3$: $1, 3, 3, 1$
• $n = 4$: $1, 4, 6, 4, 1$
• $n = 5$: $1, 5, 10, 10, 5, 1$
• $n = 6$: $1, 6, 15, 20, 15, 6, 1$

Entry $k$ of row $n$ (counting from $k = 0$) is $\binom{n}{k}$, so Pascal's triangle is just a table of binomial coefficients.

Sum identities

$\displaystyle\sum_{k = 0}^{n} \binom{n}{k} = 2^n$ (set $a = b = 1$): the sum of each row of Pascal's triangle is a power of $2$.

$\displaystyle\sum_{k = 0}^{n} (-1)^k \binom{n}{k} = 0$ for $n \ge 1$ (set $a = 1$, $b = -1$): the alternating row sum is $0$.

$\displaystyle\sum_{k = 0}^{n} k \binom{n}{k} = n \, 2^{n - 1}$ (differentiate $(1 + x)^n$ then set $x = 1$).

These show up regularly in HSC binomial-identity proofs, almost always proved by substituting convenient values of $a$ and $b$ into the binomial theorem.

Finding specific terms

Coefficient of $x^m$ in $(a x + b)^n$

write $T_{k+1} = \binom{n}{k}(ax)^{n-k} b^k$, find the $k$ that gives $x^m$, then read the coefficient.

Term independent of $x$ (the constant term)

set the power of $x$ in $T_{k + 1}$ to $0$ and solve for $k$.

Approximation $(1 + x)^n \approx 1 + n x + \binom{n}{2} x^2 + \dots$ for small $x$

keep only the first few terms, because higher powers of a small $x$ are negligible.

How exam questions ask about the binomial theorem

• "Expand $(a + b)^n$" for small $n$: read the coefficients off Pascal's triangle and attach the descending and ascending powers.
• "Find the coefficient of $x^m$ in ...": general term, solve the power equation for $k$, evaluate. The most common type.
• "Find the term independent of $x$ / the constant term in ...": general term, set the power of $x$ to $0$, solve for $k$.
• "Find the middle term of ...": for even $n$ the middle term is $T_{n/2 + 1}$.
• "Find the greatest coefficient / the greatest term when $x = \ldots$": form the ratio $\frac{T_{k+1}}{T_k}$ and find where it crosses $1$.
• "Show that $\sum \binom{n}{k} = 2^n$" (or an alternating / weighted sum): substitute suitable $a$, $b$ (often $1$ and $\pm 1$) into the binomial theorem, or differentiate $(1+x)^n$ first.
• "Use the expansion to approximate $(1.0\ldots)^n$": write the base as $1 + x$ with small $x$ and keep the first two or three terms.
• A $\left(px^a + \frac{q}{x^b}\right)^n$ form is the cue that the power of $x$ in the general term is $a(n - k) - bk$; balance it to the target power.