HSC Maths Extension 1 binomial theorem and combinatorics: deep dive (2026 guide)

Step 1. Write the general term. With $a = 2x$, $b = 5$, $n = 7$:

Step 2. Match the target power. The power of $x$ is $7 - k$. Set $7 - k = 3$, so $k = 4$.

Step 3. Substitute $k = 4$. Pick off the coefficient at this value of $k$.

Step 4. Evaluate the numerical factors. Compute $\binom{7}{4} = 35$, $2^3 = 8$, $5^4 = 625$.

The coefficient of $x^3$ in $(2x + 5)^7$ is $175\,000$.

Step 1. Write the general term. With $a = x^2$, $b = 3/x$, $n = 6$:

Step 2. Set the power of $x$ to zero. For the term to be independent of $x$, $12 - 3k = 0$, so $k = 4$.

Step 3. Substitute $k = 4$.

Step 4. Evaluate. $\binom{6}{4} = 15$ and $3^4 = 81$, so the term is $15 \cdot 81 = 1215$.

The term independent of $x$ is $1215$.

Step 1. Identify the general terms

From $(1 + 2x)^5$ the term in $x^j$ is $\binom{5}{j} 2^j x^j$. From $(1 - x)^4$ the term in $x^i$ is $\binom{4}{i} (-1)^i x^i$.

Step 2. Set up the convolution

To get $x^4$, choose $j + i = 4$ with $0 \leq j \leq 5$ and $0 \leq i \leq 4$. The valid pairs $(j, i)$ are $(0, 4)$, $(1, 3)$, $(2, 2)$, $(3, 1)$, $(4, 0)$.

Step 3. Compute each contribution

Multiply the two coefficients for each pair.

• $(0, 4)$: $\binom{5}{0} 2^0 \cdot \binom{4}{4} (-1)^4 = 1 \cdot 1 \cdot 1 \cdot 1 = 1$.
• $(1, 3)$: $\binom{5}{1} 2^1 \cdot \binom{4}{3} (-1)^3 = 5 \cdot 2 \cdot 4 \cdot (-1) = -40$.
• $(2, 2)$: $\binom{5}{2} 2^2 \cdot \binom{4}{2} (-1)^2 = 10 \cdot 4 \cdot 6 \cdot 1 = 240$.
• $(3, 1)$: $\binom{5}{3} 2^3 \cdot \binom{4}{1} (-1)^1 = 10 \cdot 8 \cdot 4 \cdot (-1) = -320$.
• $(4, 0)$: $\binom{5}{4} 2^4 \cdot \binom{4}{0} (-1)^0 = 5 \cdot 16 \cdot 1 \cdot 1 = 80$.

Step 4. Sum the contributions.

The coefficient of $x^4$ in $(1 + 2x)^5 (1 - x)^4$ is $-39$.

Step 1. Write the general term. $T_{k+1} = \binom{20}{k} 2^k$ for $k = 0, 1, \dots, 20$.

Step 2. Form the ratio of consecutive terms.

Step 3. Find where the ratio crosses $1$. Terms grow while the ratio exceeds $1$ and shrink once it drops below $1$. Solve $\frac{2(20 - k)}{k+1} > 1$:

So $T_{k+2} > T_{k+1}$ for $k < 13$ and $T_{k+2} < T_{k+1}$ for $k > 13$.

Step 4. Identify the largest term. The growth stops after $k = 13$. So the largest term is $T_{14}$, corresponding to $k = 13$.

Step 5. Verify at the boundary. At $k = 13$: ratio $= \frac{2 \cdot 7}{14} = 1$, meaning $T_{15} = T_{14}$. Both $T_{14}$ and $T_{15}$ are equal and are the largest terms. (If a single answer is required, give $k = 13$, so the greatest term is $T_{14} = \binom{20}{13} 2^{13}$.)

Step 1. Restate the question

At $x = 1$ the term in $(1+x)^{12}$ corresponding to power $r$ is $\binom{12}{r}$. We want the value of $r$ that maximises $\binom{12}{r}$ for $r \in \{0, 1, \dots, 12\}$.

Step 2. Use symmetry

$\binom{12}{r} = \binom{12}{12 - r}$, so the sequence of coefficients is symmetric about $r = 6$.

Step 3. Form the ratio of consecutive coefficients

Step 4. Find where the ratio crosses $1$. Solve $\frac{12 - r}{r + 1} > 1$:

So $\binom{12}{r+1} > \binom{12}{r}$ for $r < 5.5$, and the coefficients keep increasing until $r = 6$.

Step 5. Conclude. The greatest binomial coefficient is $\binom{12}{6} = 924$ at $r = 6$.

Step 1. Start from the binomial theorem. Set up the identity with arbitrary $a$, $b$:

Step 2. Substitute $a = 1$ and $b = 1$. This collapses the powers of $a$, $b$ to $1$.

Step 3. Simplify both sides. LHS $= 2^n$. RHS $= \sum_{k=0}^{n} \binom{n}{k}$.

Conclusion. $\sum_{k=0}^{n} \binom{n}{k} = 2^n$ for all $n \geq 0$.

Step 1. Substitute $a = 1$, $b = 1$ into the binomial theorem.

Step 2. Substitute $a = 1$, $b = -1$ into the binomial theorem. For $n \geq 1$:

Step 3. Add (i) and (ii). The odd-indexed terms cancel; the even-indexed ones double.

So $\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \dots = 2^{n-1}$.

Step 4. Subtract (ii) from (i). The even-indexed terms cancel; the odd-indexed ones double.

So $\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots = 2^{n-1}$.

Conclusion. Both the even-indexed and odd-indexed sums equal $2^{n-1}$, and so they equal each other.

Q1

Step 1. Write the general term

$T_{k+1} = \binom{8}{k} 2^{8-k} (3x)^{k} = \binom{8}{k} 2^{8-k} 3^k x^k$.

Step 2. Match the target power

Set $k = 5$.

Step 3. Substitute

$T_6 = \binom{8}{5} 2^3 3^5 x^5 = 56 \cdot 8 \cdot 243 \, x^5$.

Step 4. Evaluate

$56 \cdot 8 = 448$ and $448 \cdot 243 = 108\,864$.

The coefficient of $x^5$ is $108\,864$.

Q2

Step 1. Write the general term

$T_{k+1} = \binom{9}{k} (2x)^{9-k} \left( -\dfrac{1}{x^2} \right)^k = \binom{9}{k} 2^{9-k} (-1)^k \, x^{9-k} \cdot x^{-2k} = \binom{9}{k} 2^{9-k} (-1)^k \, x^{9 - 3k}$.

Step 2. Set the power of $x$ to zero

$9 - 3k = 0$, so $k = 3$.

Step 3. Substitute $k = 3$

$T_4 = \binom{9}{3} 2^{6} (-1)^3 = 84 \cdot 64 \cdot (-1)$.

Step 4. Evaluate

$84 \cdot 64 = 5376$, so the term is $-5376$.

The term independent of $x$ is $-5376$.

Q3

Step 1. Start from the binomial theorem

$(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k$.

Step 2. Substitute $x = 2$

$(1 + 2)^n = \sum_{k=0}^{n} \binom{n}{k} 2^k$.

Step 3. Simplify

LHS $= 3^n$.

Conclusion

$\sum_{k=0}^{n} \binom{n}{k} 2^k = 3^n$ for all integers $n \geq 0$.

Q4

Step 1. Identify the general terms

From $(1+x)^5$, the term in $x^j$ is $\binom{5}{j} x^j$. From $(1+2x)^4$, the term in $x^i$ is $\binom{4}{i} 2^i x^i$.

Step 2. List the pairs giving $x^2$

$(j, i) \in \{(0, 2), (1, 1), (2, 0)\}$.

Step 3. Compute contributions

• $(0, 2)$: $\binom{5}{0} \cdot \binom{4}{2} 2^2 = 1 \cdot 6 \cdot 4 = 24$.
• $(1, 1)$: $\binom{5}{1} \cdot \binom{4}{1} 2^1 = 5 \cdot 4 \cdot 2 = 40$.
• $(2, 0)$: $\binom{5}{2} \cdot \binom{4}{0} = 10 \cdot 1 = 10$.

Step 4. Sum. $24 + 40 + 10 = 74$.

The coefficient of $x^2$ is $74$.

Q5

Step 1. Use the ratio of consecutive coefficients

$\dfrac{\binom{15}{r+1}}{\binom{15}{r}} = \dfrac{15 - r}{r + 1}$.

Step 2. Solve ratio $> 1$

$15 - r > r + 1 \implies r < 7$.

Step 3. Interpret

Coefficients increase while $r < 7$, so the largest is at $r = 7$ or $r = 8$. Check: $\binom{15}{7} = 6435 = \binom{15}{8}$. They are equal.

The greatest binomial coefficient is $\binom{15}{7} = \binom{15}{8} = 6435$.

Q6

Step 1. Identify the distinct letters available

MATHEMATICS contains the distinct letters $\{M, A, T, H, E, I, C, S\}$, that is $8$ different letters. The vowels are $A, E, I$ ($3$ vowels) and the consonants are $M, T, H, C, S$ ($5$ consonants).

Step 2. Count total selections of $4$ distinct letters

$\binom{8}{4} = 70$.

Step 3. Count selections with no vowels (complement)

Choose all $4$ from the $5$ consonants: $\binom{5}{4} = 5$.

Step 4. Apply complementary counting

Selections with at least one vowel $= 70 - 5 = 65$.

The number of selections is $65$.

Q7

Step 1. Write the general term at $x = 1$

$T_{k+1} = \binom{18}{k} 3^k$ for $k = 0, 1, \dots, 18$.

Step 2. Form the ratio of consecutive terms

Step 3. Solve ratio $> 1$. $3(18 - k) > k + 1 \implies 54 - 3k > k + 1 \implies 53 > 4k \implies k < 13.25$.

Step 4. Interpret. Terms increase while $k < 13.25$, so the largest is at $k = 13$ (giving $T_{14}$).

The greatest term is $T_{14} = \binom{18}{13} 3^{13}$, corresponding to $k = 13$.

Q8

Step 1. Write the binomial expansion

$(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k$.

Step 2. Differentiate both sides with respect to $x$

Use the chain rule on the LHS and term-by-term differentiation on the RHS.

Step 3. Substitute $x = 1$. LHS $= n \cdot 2^{n-1}$. RHS $= \sum_{k=0}^{n} k \binom{n}{k}$.

Conclusion. $\sum_{k=0}^{n} k \binom{n}{k} = n \cdot 2^{n-1}$ for all integers $n \geq 1$.