Solved Examples

How to Use This Section

Thirty fully worked examples sweep the whole chapter — Pascal-row expansions, alternating signs, fractional terms, coefficient extraction by inspection, numerical evaluations and approximations, comparisons, divisibility proofs, conjugate-pair simplifications, coefficient-identity drills, and full trinomial expansions solved by grouping.

[Board Tip] Two sanity checks after every expansion: the term count must be $n + 1$, and substituting $x = 1$ must reproduce the value of the original expression at $x = 1$. Ten seconds, catches most slips.

Expansions

Example 1: A clean quartic

Expand $(2x + 3)^4$.

Solution:

Step 1 — Fetch row 4. $1, 4, 6, 4, 1$.

Step 2 — Attach the powers. $(2x)^4 + 4(2x)^3(3) + 6(2x)^2(9) + 4(2x)(27) + 81$.

Step 3 — Evaluate. $16x^4 + 96x^3 + 216x^2 + 216x + 81$.

Step 4 — Check. At $x = 1$: $16 + 96 + 216 + 216 + 81 = 625 = 5^4 = (2+3)^4$ ✓.

Takeaway: Five terms for index 4, and the $x=1$ check confirms the whole line.

Example 2: Alternating signs, constant first

Expand $(3 - 2x)^5$.

Solution:

Step 1 — Row 5 with alternating signs. $3^5 - 5(3^4)(2x) + 10(3^3)(2x)^2 - 10(3^2)(2x)^3 + 5(3)(2x)^4 - (2x)^5$.

Step 2 — Evaluate each coefficient. $243, \ 5 \cdot 81 \cdot 2 = 810, \ 10 \cdot 27 \cdot 4 = 1080, \ 10 \cdot 9 \cdot 8 = 720, \ 5 \cdot 3 \cdot 16 = 240, \ 32$.

Step 3 — Assemble. $243 - 810x + 1080x^2 - 720x^3 + 240x^4 - 32x^5$.

Step 4 — Check. At $x = 1$: $243 - 810 + 1080 - 720 + 240 - 32 = 1 = (3-2)^5$ ✓.

Takeaway: The constant leads when the variable sits second — the signs still alternate off the minus.

Example 3: Two variables with a coefficient

Expand $(2a - 3b)^4$.

Solution:

Step 1 — Row 4 with alternating signs. $(2a)^4 - 4(2a)^3(3b) + 6(2a)^2(9b^2) - 4(2a)(27b^3) + 81b^4$.

Step 2 — Evaluate. $16a^4 - 96a^3b + 216a^2b^2 - 216ab^3 + 81b^4$.

Step 3 — Check. At $a = b = 1$: $16 - 96 + 216 - 216 + 81 = 1 = (2-3)^4$ ✓.

Takeaway: Both quantities carry coefficients — bracket each before raising.

Example 4: Coefficient by inspection

Find the coefficient of $x^6y^3$ in $(x^2 - y)^6$.

Solution:

Step 1 — Locate the term. $y^3$ requires the $r = 3$ term: $^6C_3(x^2)^3(-y)^3$.

Step 2 — Confirm the $x$-power. $(x^2)^3 = x^6$ ✓ — the term is exactly $x^6y^3$-shaped.

Step 3 — Evaluate the sign and number. $20 \cdot x^6 \cdot (-y^3)$: coefficient $-20$.

Takeaway: Write one general term, match the powers, then evaluate — no full expansion needed.

Example 5: Coefficient of $a^2b^3$

Find the coefficient of $a^2b^3$ in $(a + 2b)^5$.

Solution:

Step 1 — Locate. The $b^3$ term: $^5C_3\,a^2(2b)^3$.

Step 2 — Evaluate. $10 \times 8 = 80$.

Takeaway: The inner coefficient 2 contributes $2^3 = 8$ — never just 2.

Example 6: Powers of $x$ in a mixed expansion

Find the coefficient of $x$ in $\left(x^2 + \frac{2}{x}\right)^5$.

Solution:

Step 1 — Write the general term. $^5C_r(x^2)^{5-r}\left(\frac{2}{x}\right)^r = {^5C_r}\,2^r\,x^{10-3r}$.

Step 2 — Solve for the power. $10 - 3r = 1$ gives $r = 3$.

Step 3 — Evaluate. $^5C_3\,2^3 = 10 \times 8 = 80$.

Takeaway: Collapse the powers of $x$ into one exponent formula, then solve for $r$.

Example 7: A trinomial by grouping

Expand $\left(1 + \frac{x}{2} - \frac{2}{x}\right)^4$, $x \ne 0$.

Solution:

Step 1 — Group into a binomial. Let $u = \frac{x}{2} - \frac{2}{x}$; expand $(1 + u)^4 = 1 + 4u + 6u^2 + 4u^3 + u^4$.

Step 2 — Expand the powers of $u$. $u^2 = \frac{x^2}{4} - 2 + \frac{4}{x^2}$; $u^3 = \frac{x^3}{8} - \frac{3x}{2} + \frac{6}{x} - \frac{8}{x^3}$; $u^4 = \frac{x^4}{16} - x^2 + 6 - \frac{16}{x^2} + \frac{16}{x^4}$.

Step 3 — Substitute and collect like powers.

$$\frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} - 4x - 5 + \frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4}$$

Step 4 — Check the constant. $1 + 6(-2) + 6 = -5$ ✓ (from $1, 6u^2, u^4$).

Takeaway: Trinomials fold into binomials by grouping — then it is bookkeeping, done power by power.

Example 8: A squared-trinomial cube

Expand $(3x^2 - 2ax + 3a^2)^3$ using the binomial theorem.

Solution:

Step 1 — Group. $\left[(3x^2 - 2ax) + 3a^2\right]^3 = A^3 + 3A^2(3a^2) + 3A(9a^4) + 27a^6$ with $A = 3x^2 - 2ax$.

Step 2 — Expand the powers of $A$. $A^2 = 9x^4 - 12ax^3 + 4a^2x^2$; $A^3 = 27x^6 - 54ax^5 + 36a^2x^4 - 8a^3x^3$.

Step 3 — Collect.

$$27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6$$

Step 4 — Check. At $x = a = 1$: $27 - 54 + 117 - 116 + 117 - 54 + 27 = 64 = (3 - 2 + 3)^3$ ✓.

Takeaway: The coefficients read the same forwards and backwards — the trinomial is symmetric in $x$ and $a$.

Numerical Evaluations and Comparisons

Example 9: A fourth power near 100

Evaluate $(99)^4$.

Solution:

Step 1 — Split. $(100 - 1)^4$ with row $1, 4, 6, 4, 1$.

Step 2 — Write the terms. $10^8 - 4(10^6) + 6(10^4) - 4(100) + 1$.

Step 3 — Add. $100000000 - 4000000 + 60000 - 400 + 1 = 96059601$.

Takeaway: A $\pm 1$ split leaves only the row entries and decimal shifts.

Example 10: Decimal base

Approximate $(1.02)^8$ using three terms.

Solution:

Step 1 — Split and truncate. $(1 + 0.02)^8 \approx 1 + 8(0.02) + {^8C_2}(0.02)^2$.

Step 2 — Evaluate. $1 + 0.16 + 28(0.0004) = 1 + 0.16 + 0.0112 = 1.1712$.

Step 3 — Estimate the error. Next term $^8C_3(0.02)^3 = 56 \times 8 \times 10^{-6} \approx 0.0004$ — the true value is $1.1717\ldots$

Takeaway: Three terms of a small-$x$ expansion already land within $10^{-3}$.

Example 11: Three-term comparison

Which is larger: $(1.2)^{10}$ or $4$?

Solution:

Step 1 — Try two terms. $1 + 10(0.2) = 3$ — smaller than 4, not yet conclusive.

Step 2 — Add the third term. $^{10}C_2(0.2)^2 = 45 \times 0.04 = 1.8$: the partial sum is $1 + 2 + 1.8 = 4.8$.

Step 3 — Conclude. $(1.2)^{10} > 4.8 > 4$ — larger.

Takeaway: If two terms don't settle a comparison, keep adding positive terms until they do.

Example 12: A power of 1.1, quickly

Approximate $(1.1)^5$ with three terms.

Solution:

Step 1 — Truncate. $1 + 5(0.1) + 10(0.01) = 1 + 0.5 + 0.1 = 1.6$.

Step 2 — Gauge the error. Next term $10(0.1)^3 = 0.01$; true value $1.61051$.

Takeaway: The first dropped term estimates the error — quote it to justify the cut.

Example 13: Divisibility by 49

Prove that $2^{3n} - 7n - 1$ is divisible by 49 for every positive integer $n$.

Solution:

Step 1 — Rewrite the power. $2^{3n} = 8^n = (1 + 7)^n$.

Step 2 — Expand and group. $(1+7)^n = 1 + 7n + 49\,{^nC_2} + 343\,{^nC_3} + \cdots$

Step 3 — Subtract. $2^{3n} - 7n - 1 = 49\left({^nC_2} + 7\,{^nC_3} + \cdots\right)$ — a multiple of 49. ∎

Takeaway: Recognise $2^{3n}$ as $8^n$ first — the split $1 + 7$ then produces the 49's.

Example 14: Divisibility by 9

Show $4^n - 3n - 1$ is divisible by 9 for all positive integers $n$.

Solution:

Step 1 — Split. $4^n = (1+3)^n = 1 + 3n + 9\,{^nC_2} + 27\,{^nC_3} + \cdots$

Step 2 — Subtract. $4^n - 3n - 1 = 9\left({^nC_2} + 3\,{^nC_3} + \cdots\right)$. ∎

Takeaway: The modulus 9 is $3^2$ — split off 3's and the third term onward delivers it.

Example 15: Remainder, then subtraction

What is the remainder when $6^n - 5n - 1$ is divided by 25?

Solution:

Step 1 — Start from the mod-25 result. $6^n - 5n = 25k + 1$ (from the split $6 = 1 + 5$).

Step 2 — Subtract 1. $6^n - 5n - 1 = 25k$.

Step 3 — Conclude. Remainder $0$ — the expression is exactly divisible by 25.

Takeaway: Once the remainder identity is proved, nearby questions are one-line corollaries.

Conjugate Pairs and Surds

Example 16: A sixth-power conjugate difference

Evaluate $(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6$.

Solution:

Step 1 — Keep the odd-power terms, doubled. With $a = \sqrt{3}, b = \sqrt{2}$:

$$2\left[{^6C_1}a^5b + {^6C_3}a^3b^3 + {^6C_5}ab^5\right]$$

Step 2 — Evaluate the surd powers. $a^5 = 9\sqrt{3}$, $a^3 = 3\sqrt{3}$, $b^3 = 2\sqrt{2}$, $b^5 = 4\sqrt{2}$.

Step 3 — Compute each term. $6 \cdot 9\sqrt{3} \cdot \sqrt{2} = 54\sqrt{6}$; $20 \cdot 3\sqrt{3} \cdot 2\sqrt{2} = 120\sqrt{6}$; $6 \cdot \sqrt{3} \cdot 4\sqrt{2} = 24\sqrt{6}$.

Step 4 — Add and double. $2(54 + 120 + 24)\sqrt{6} = 396\sqrt{6}$.

Takeaway: Every surviving term carries $\sqrt{3}\sqrt{2} = \sqrt{6}$ to an odd power — the answer collects on $\sqrt{6}$.

Example 17: A parametric conjugate sum

Find the value of $\left(a^2 + \sqrt{a^2 - 1}\right)^4 + \left(a^2 - \sqrt{a^2 - 1}\right)^4$.

Solution:

Step 1 — Conjugate sum keeps even powers. With $A = a^2$, $B = \sqrt{a^2-1}$:

$$2\left[A^4 + {^4C_2}A^2B^2 + B^4\right] = 2\left[A^4 + 6A^2B^2 + B^4\right]$$

Step 2 — Substitute $B^2 = a^2 - 1$. $2\left[a^8 + 6a^4(a^2-1) + (a^2-1)^2\right]$.

Step 3 — Expand and collect. $2\left[a^8 + 6a^6 - 6a^4 + a^4 - 2a^2 + 1\right] = 2a^8 + 12a^6 - 10a^4 - 4a^2 + 2$.

Takeaway: Even powers of $B$ are polynomials in $a$ — the square root disappears without ever being evaluated.

Example 18: A fifth-power conjugate sum

Evaluate $(\sqrt{2} + 1)^5 + (\sqrt{2} - 1)^5$.

Solution:

Step 1 — Keep even powers of $b = 1$. $2\left[{^5C_0}a^5 + {^5C_2}a^3 + {^5C_4}a\right]$ with $a = \sqrt{2}$.

Step 2 — Evaluate the surd powers. $a^5 = 4\sqrt{2}$, $a^3 = 2\sqrt{2}$: bracket $= 4\sqrt{2} + 10 \cdot 2\sqrt{2} + 5\sqrt{2} = 29\sqrt{2}$.

Step 3 — Double. $2 \times 29\sqrt{2} = 58\sqrt{2}$.

Takeaway: With an ODD index the even-power terms hold odd powers of $a$ — the surd survives, tidily collected.

Example 19: Fourth-power conjugate sum

Evaluate $(\sqrt{3} + 1)^4 + (\sqrt{3} - 1)^4$.

Solution:

Step 1 — Even powers, doubled. $2\left[{^4C_0}a^4 + {^4C_2}a^2 + {^4C_4}\right]$, $a = \sqrt{3}$.

Step 2 — Evaluate. $a^4 = 9$, $a^2 = 3$: $2\left[9 + 18 + 1\right]$.

Step 3 — Compute. $2 \times 28 = 56$.

Takeaway: Even index + conjugate sum = an integer every time — the surd appears only in even powers.

Example 20: Is it an integer?

Show that $(2 + \sqrt{3})^4 + (2 - \sqrt{3})^4$ is an integer and find it.

Solution:

Step 1 — Conjugate sum. $2\left[2^4 + {^4C_2} \cdot 2^2 \cdot 3 + 3^2\right]$ — only even powers of $\sqrt{3}$, all rational.

Step 2 — Evaluate. $2\left[16 + 72 + 9\right] = 2 \times 97 = 194$.

Takeaway: Conjugate surd pairs always sum to integers — the odd (irrational) terms cancel by construction.

Coefficient Identities and Structure

Example 21: Sum of all coefficients

Find the sum of the coefficients in the expansion of $(1 + 2x)^4$.

Solution:

Step 1 — Substitute $x = 1$. Every term becomes its bare coefficient: $(1 + 2)^4 = 81$.

Step 2 — Verify against the expansion. $1 + 8x + 24x^2 + 32x^3 + 16x^4$: $1 + 8 + 24 + 32 + 16 = 81$ ✓.

Takeaway: "Sum of coefficients" always means: substitute 1 for every variable.

Example 22: A sum that collapses to $-1$

Find the sum of the coefficients of $(2x - 3y)^9$.

Solution:

Step 1 — Substitute $x = y = 1$. $(2 - 3)^9 = (-1)^9$.

Step 2 — Evaluate. $-1$ — no expansion required.

Takeaway: The substitution works for any number of variables, even when the base goes negative.

Example 23: Identifying $n$ from the coefficient sum

If the sum of all binomial coefficients of $(1 + x)^n$ is 512, find $n$.

Solution:

Step 1 — Use the identity. $\sum {^nC_r} = 2^n = 512$.

Step 2 — Solve. $512 = 2^9$, so $n = 9$.

Takeaway: Coefficient-sum data converts directly to a power of 2.

Example 24: Even-position sum

Prove $^nC_0 + {^nC_2} + {^nC_4} + \cdots = 2^{n-1}$.

Solution:

Step 1 — Write the two identities. At $x = 1$: $\sum {^nC_r} = 2^n$; at $x = -1$: $\sum (-1)^r\,{^nC_r} = 0$.

Step 2 — Add them. Odd-position terms cancel; even ones double: $2\left({^nC_0} + {^nC_2} + \cdots\right) = 2^n$.

Step 3 — Divide. Even-position sum $= 2^{n-1}$. ∎ (Subtracting instead gives the odd-position sum, also $2^{n-1}$.)

Takeaway: Add the identities for the even split, subtract for the odd one.

Example 25: Equal coefficients locate $n$

In $(1 + x)^n$ the coefficients of $x^7$ and $x^8$ are equal. Find $n$.

Solution:

Step 1 — Set up. $^nC_7 = {^nC_8}$.

Step 2 — Apply the equality rule. $7 \ne 8$, so $n = 7 + 8 = 15$.

Step 3 — Check. $^{15}C_7 = {^{15}C_8} = 6435$ ✓ by symmetry.

Takeaway: Equal binomial coefficients with different lower indices always mean the indices add to $n$.

Example 26: Term count

The expansion of $(a + b)^n$ has 12 terms. Find $n$, and the largest binomial coefficient for that index.

Solution:

Step 1 — Term count. $n + 1 = 12$, so $n = 11$.

Step 2 — Locate the peak. For odd $n$ there are TWO equal largest coefficients, at the middle: $^{11}C_5 = {^{11}C_6}$.

Step 3 — Evaluate. $^{11}C_5 = 462$.

Takeaway: Odd index → twin peaks; even index → single central peak.

Example 27: The largest coefficient in an even-index row

Which coefficient of $(1 + x)^8$ is largest?

Solution:

Step 1 — Peak of an even row. Row 8 climbs to its centre entry $^8C_4$.

Step 2 — Evaluate. $^8C_4 = 70$, the coefficient of $x^4$.

Takeaway: Binomial rows increase to the middle and mirror back down.

Example 28: A parameter from equal coefficients

If the coefficients of $x^2$ and $x^3$ in $(3 + ax)^9$ are equal, find $a$.

Solution:

Step 1 — Write both coefficients. Of $x^2$: $^9C_2\,3^7a^2 = 36 \cdot 3^7 a^2$; of $x^3$: $^9C_3\,3^6a^3 = 84 \cdot 3^6 a^3$.

Step 2 — Equate and cancel. $36 \cdot 3^7 a^2 = 84 \cdot 3^6 a^3$; divide by $3^6 a^2$: $36 \times 3 = 84a$.

Step 3 — Solve. $a = \frac{108}{84} = \frac{9}{7}$.

Takeaway: Cancel the common powers before solving — the equation collapses to one line.

Example 29: Reading a sigma sum

Evaluate $\sum_{r=0}^{8} {^8C_r}\,2^r$.

Solution:

Step 1 — Pattern-match. $\sum {^nC_r}t^r = (1+t)^n$ with $t = 2$, $n = 8$.

Step 2 — Evaluate. $(1 + 2)^8 = 3^8 = 6561$.

Takeaway: Weighted coefficient sums are expansions read backwards.

Example 30: An alternating weighted sum

Evaluate $\sum_{r=0}^{6} (-1)^r\,{^6C_r}\,2^{6-r}$.

Solution:

Step 1 — Pattern-match. $\sum {^6C_r}\,2^{6-r}(-1)^r = (2 + (-1))^6$.

Step 2 — Evaluate. $(2-1)^6 = 1$.

Takeaway: Falling powers of 2 and alternating signs are just $(a+b)^n$ with $a = 2$, $b = -1$.