Binomial Theorem Expansion

The Binomial Theorem

The binomial theorem expands $(a + b)^n$ without multiplying it out the long way:

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

The coefficients $\binom{n}{k}$ come from Pascal's triangle or the formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Step-by-step: Expand $(2x + 3)^5$

Step 1 — Identify $a = 2x$, $b = 3$, $n = 5$.

Step 2 — Get Pascal's row 5: $1, 5, 10, 10, 5, 1$.

Step 3 — Write each term $\binom{5}{k}(2x)^{5-k}(3)^k$:

• $k=0$: $1 \cdot (2x)^5 \cdot 3^0 = 1 \cdot 32x^5 \cdot 1 = 32x^5$
• $k=1$: $5 \cdot (2x)^4 \cdot 3^1 = 5 \cdot 16x^4 \cdot 3 = 240x^4$
• $k=2$: $10 \cdot (2x)^3 \cdot 3^2 = 10 \cdot 8x^3 \cdot 9 = 720x^3$
• $k=3$: $10 \cdot (2x)^2 \cdot 3^3 = 10 \cdot 4x^2 \cdot 27 = 1080x^2$
• $k=4$: $5 \cdot (2x)^1 \cdot 3^4 = 5 \cdot 2x \cdot 81 = 810x$
• $k=5$: $1 \cdot (2x)^0 \cdot 3^5 = 1 \cdot 1 \cdot 243 = 243$

Step 4 — Combine:

$(2x + 3)^5 = 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243$

Step 5 — Quick check: Set $x = 0$: $(3)^5 = 243$. Our expansion gives $243$ ✓.

Set $x = 1$: $(5)^5 = 3125$. Sum: $32 + 240 + 720 + 1080 + 810 + 243 = 3125$ ✓.

Common mistakes

• Forgetting to raise $a$ to a power. Each term has $(2x)^{5-k}$, NOT just $x^{5-k}$. The coefficient 2 gets raised to a power too: $(2x)^4 = 16x^4$, not $2x^4$.
• Using wrong Pascal's row. $(a+b)^5$ uses row 5 (6 numbers: 1,5,10,10,5,1), not row 4.

Try it in the visualization

Adjust $a$, $b$, $n$. Pascal's triangle highlights the relevant row. Each term is computed with the coefficient bar chart showing their relative sizes.