Binomial theorem for positive integral indices

You already know how to expand $(a + b)^2 = a^2 + 2ab + b^2$ and $(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$. The binomial theorem gives the formula for $(a + b)^n$ for any positive integer $n$ , all in one stroke.

The theorem

Theorem. For any positive integer $n$ and any real (or complex) $a, b$: $\boxed{(a + b)^n = \sum_{r = 0}^{n} \binom{n}{r}\,a^{n - r}\,b^r}$ where $\binom{n}{r} = \dfrac{n!}{r!\,(n - r)!}$.

Written out term by term: $(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n-1} b + \binom{n}{2} a^{n-2} b^2 + \dots + \binom{n}{n-1} a b^{n-1} + \binom{n}{n} b^n.$

There are $n + 1$ terms. The first $\binom{n}{0} = 1$ and the last $\binom{n}{n} = 1$. The coefficients form the $n$-th row of Pascal's triangle.

Combinatorial proof

When we expand $(a + b)^n = (a + b)(a + b) \cdots (a + b)$ ($n$ factors), each term in the expansion is formed by choosing either $a$ or $b$ from each of the $n$ factors and multiplying. A term that uses $b$ from exactly $r$ factors and $a$ from the other $n - r$ contributes $a^{n - r} b^r$. The number of ways to choose those $r$ factors is $\binom{n}{r}$. Hence $(a + b)^n = \sum_{r=0}^{n} \binom{n}{r}\,a^{n - r}\,b^r. \qed$

Proof by induction

Base. $n = 1$: $(a + b)^1 = a + b = \binom{1}{0} a + \binom{1}{1} b$. ✓

Step. Assume true for some $n$. Then $(a + b)^{n + 1} = (a + b) \cdot (a + b)^n = (a + b) \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r.$

Distribute: $= \sum_{r=0}^n \binom{n}{r} a^{n+1-r} b^r + \sum_{r=0}^n \binom{n}{r} a^{n-r} b^{r+1}.$

Reindex the second sum (let $s = r + 1$): $= \sum_{r=0}^n \binom{n}{r} a^{n+1-r} b^r + \sum_{s=1}^{n+1} \binom{n}{s-1} a^{n+1-s} b^s.$

Combine like terms (in terms of $r$, $s$ being just dummy variables): $= a^{n+1} + b^{n+1} + \sum_{r=1}^{n} \left[\binom{n}{r} + \binom{n}{r - 1}\right] a^{n+1-r} b^r.$

By Pascal's identity, $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$. So $(a + b)^{n+1} = \sum_{r=0}^{n+1} \binom{n+1}{r} a^{n+1-r} b^r. \qed$

Variations and special cases

Setting $a = 1, b = x$: $(1 + x)^n = \sum_{r=0}^n \binom{n}{r} x^r.$

Setting $a = 1, b = -x$: $(1 - x)^n = \sum_{r=0}^n (-1)^r \binom{n}{r} x^r.$

Sums of coefficients (set $x = 1$): $\sum_{r=0}^n \binom{n}{r} = 2^n.$

Setting $x = -1$: $\sum_{r=0}^n (-1)^r \binom{n}{r} = 0.$

Worked examples

Example 1. Expand $(x + 2)^4$.

$\binom{4}{0} x^4 + \binom{4}{1} x^3 \cdot 2 + \binom{4}{2} x^2 \cdot 4 + \binom{4}{3} x \cdot 8 + \binom{4}{4} \cdot 16$ $= x^4 + 8 x^3 + 24 x^2 + 32 x + 16$.

Example 2. Expand $(2x - 3y)^3$.

$\binom{3}{0} (2x)^3 + \binom{3}{1} (2x)^2 (-3y) + \binom{3}{2} (2x)(-3y)^2 + \binom{3}{3} (-3y)^3$ $= 8 x^3 - 36 x^2 y + 54 x y^2 - 27 y^3$.

Example 3. Use the binomial theorem to compute $(1.01)^5$ approximately.

$(1 + 0.01)^5 = 1 + 5(0.01) + 10(0.01)^2 + 10(0.01)^3 + 5(0.01)^4 + (0.01)^5$ $= 1 + 0.05 + 0.001 + \dots \approx 1.05101$.

Example 4. Show $\sum_{r=0}^n \binom{n}{r} = 2^n$.

Set $a = b = 1$ in the binomial theorem: $2^n = (1 + 1)^n = \sum_{r=0}^n \binom{n}{r}. \qed$

Example 5 (harder). Prove $\sum_{r=0}^n r \binom{n}{r} = n \cdot 2^{n-1}$.

Use the identity $r \binom{n}{r} = n \binom{n-1}{r-1}$: $\sum_{r=0}^n r \binom{n}{r} = \sum_{r=1}^n n \binom{n-1}{r-1} = n \sum_{s=0}^{n-1} \binom{n-1}{s} = n \cdot 2^{n-1}. \qed$

Alternatively, differentiate $(1 + x)^n = \sum \binom{n}{r} x^r$ with respect to $x$: $n(1 + x)^{n-1} = \sum r \binom{n}{r} x^{r-1}.$ Setting $x = 1$ gives the result.

Try it yourself

1. Expand $(x + 3)^4$.
2. Expand $(2x - 1)^5$.
3. Find the value of $(\sqrt{2} + 1)^4 + (\sqrt{2} - 1)^4$.
4. Compute $(99)^4$ using $99 = 100 - 1$.
5. Prove $\sum_{r=0}^n (-1)^r \binom{n}{r} = 0$.
6. Show $\binom{n}{0}^2 + \binom{n}{1}^2 + \dots + \binom{n}{n}^2 = \binom{2n}{n}$. (Hint: compare $(1 + x)^n (1 + x)^n$ to $(1 + x)^{2n}$.)
7. Expand $\left(x + \dfrac{1}{x}\right)^4$ and identify all five terms.
8. Show that the coefficient of $x^r$ in $(1 + x)^n$ equals the coefficient of $x^{n-r}$.
9. Find an approximation to $(1.02)^6$ using the first four terms.
10. Show $(a + b)^n + (a - b)^n = 2 [\binom{n}{0} a^n + \binom{n}{2} a^{n-2} b^2 + \dots]$.
11. Expand $(x^2 - 2y)^4$.
12. Prove the formula by induction in detail (write out the $n = 2$ to $n = 3$ step in full).

Pitfalls / Tricks

• The expansion of $(a + b)^n$ has $n + 1$ terms (not $n$).
• For $(a - b)^n$, signs alternate: $+, -, +, -, \dots$.
• Always identify $a, b, n$ explicitly before writing $T_{r+1}$.
• Insight. The binomial theorem is a bridge between algebra (multiplication) and combinatorics (counting). Once you see $(a + b)^n$ as "every way to pick $b$'s from $n$ factors", the formula is unforgettable.