Binomial Expansion

Board	Paper	Notes
AQA	Paper 1, 2	Binomial theorem for positive integer $n$; P2: general expansion
Edexcel	P1, P2	P1: positive integer; P2: negative/fractional
OCR (A)	Paper 1, 2	Similar split
CIE (9709)	P1, P3	P1: positive integer; P3: general

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Pascal’s triangle displays the binomial coefficients for successive powers of $(a + b)^n$:

$$\begin{array}{c} N = 0: \quad 1 \\ N = 1: \quad 1 \quad 1 \\ N = 2: \quad 1 \quad 2 \quad 1 \\ N = 3: \quad 1 \quad 3 \quad 3 \quad 1 \\ N = 4: \quad 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ \end{array}$$

Each entry is the sum of the two entries above it.

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Definition. The binomial coefficient $\binom{n}{r}$ (read ”$n$ choose $r$”) is defined for Non-negative integers $n, r$ with $r \leq n$ by:

$\binom{n}{r} = \frac{n!}{r!(n - r)!}$

Where $n! = n(n-1)(n-2)\cdots 1$ is the factorial of $n$And $0! = 1$.

Theorem. $\binom{n}{r}$ counts the number of ways to choose $r$ objects from $n$ distinct Objects (order does not matter).

Proof. We count the number of ways to form a subset of size $r$ from $\{1, 2, \ldots, n\}$.

The number of ordered selections of $r$ objects from $n$ is $n(n-1)\cdots(n-r+1) = \frac{n!}{(n-r)!}$.

But each subset of size $r$ can be ordered in $r!$ ways. Dividing by $r!$ (to account for Overcounting):

$\binom{n}{r} = \frac{n!}{r!(n-r)!} \quad \blacksquare$

Theorem (Symmetry). $\binom{n}{r} = \binom{n}{n - r}$.

Proof. $\binom{n}{n-r} = \frac{n!}{(n-r)!(n - (n-r))!} = \frac{n!}{(n-r)!\,r!} = \binom{n}{r}$. $\blacksquare$

Intuition. Choosing $r$ objects to include is equivalent to choosing $n - r$ objects to exclude.

Theorem (Pascal’s Identity). $\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$.

Proof. Consider a set of $n$ elements, with one distinguished element $x$. To choose $r$ elements:

• Case 1: Include $x$. Choose $r - 1$ more from the remaining $n - 1$: $\binom{n-1}{r-1}$ ways.
• Case 2: Exclude $x$. Choose all $r$ from the remaining $n - 1$: $\binom{n-1}{r}$ ways.

Total: $\binom{n-1}{r-1} + \binom{n-1}{r}$. $\blacksquare$

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Theorem (Binomial Theorem for Positive Integer $n$). For $n \in \mathbb{N}$:

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

$= a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n-1}ab^{n-1} + b^n$

Proof (by induction on $n$).

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

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

Inductive step: Assume $(a+b)^k = \sum_{r=0}^{k} \binom{k}{r} a^{k-r} b^r$.

$$\begin{aligned} (a+b)^{k+1} &= (a+b)(a+b)^k \\ &= (a+b)\sum_{r=0}^{k} \binom{k}{r} a^{k-r} b^r \\ &= \sum_{r=0}^{k} \binom{k}{r} a^{k+1-r} b^r + \sum_{r=0}^{k} \binom{k}{r} a^{k-r} b^{r+1} \\ &= a^{k+1} + \sum_{r=1}^{k} \binom{k}{r} a^{k+1-r} b^r + \sum_{r=1}^{k+1} \binom{k}{r-1} a^{k+1-r} b^r + b^{k+1} \\ &= a^{k+1} + \sum_{r=1}^{k} \left[\binom{k}{r} + \binom{k}{r-1}\right] a^{k+1-r} b^r + b^{k+1} \end{aligned}$$

By Pascal’s identity, $\binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r}$:

$= a^{k+1} + \sum_{r=1}^{k} \binom{k+1}{r} a^{k+1-r} b^r + b^{k+1} = \sum_{r=0}^{k+1} \binom{k+1}{r} a^{k+1-r} b^r \quad \blacksquare$

Intuition. Each term in the expansion corresponds to a way of choosing $a$‘s and $b$‘s: from $n$ Factors of $(a + b)$Choosing $r$ of them to contribute a $b$ (and the remaining $n - r$ to Contribute an $a$) gives the term $\binom{n}{r} a^{n-r} b^r$.

Example

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

$(2x - 3)^5 = \sum_{r=0}^{5} \binom{5}{r} (2x)^{5-r}(-3)^r$

$= 32x^5 + 5 \cdot 16x^4(-3) + 10 \cdot 8x^3 \cdot 9 + 10 \cdot 4x^2(-27) + 5 \cdot 2x \cdot 81 + (-243)$

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

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When $n$ is not a positive integer (e.g., $n$ is negative or fractional), the expansion becomes an infinite series.

Theorem (General Binomial Theorem). For $|x| \lt 1$ and any $n \in \mathbb{R}$:

$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$

$= \sum_{r=0}^{\infty} \binom{n}{r} x^r$

Where $\binom{n}{r} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}$.

Note. When $n$ is a positive integer, the series terminates at $r = n$ (since $\binom{n}{r} = 0$ For $r \gt n$). When $n$ is not a positive integer, the series is infinite.

Convergence. The series converges when $|x| \lt 1$. This condition is essential.

Example

Find the expansion of $(1 - 2x)^{-3}$ up to and including the term in $x^3$.

Here $n = -3$ and we replace $x$ with $-2x$.

$$\begin{aligned} (1 - 2x)^{-3} &= 1 + (-3)(-2x) + \frac{(-3)(-4)}{2!}(-2x)^2 + \frac{(-3)(-4)(-5)}{3!}(-2x)^3 + \cdots \\ &= 1 + 6x + \frac{12}{2}(4x^2) + \frac{-60}{6}(-8x^3) + \cdots \\ &= 1 + 6x + 24x^2 + 80x^3 + \cdots \end{aligned}$$

Valid for $|-2x| \lt 1$I.e., $|x| \lt \frac{1}{2}$.

To expand $(a + bx)^n$First factor out $a$:

$(a + bx)^n = a^n\left(1 + \frac{b}{a}x\right)^n$

Then expand $\left(1 + \frac{b}{a}x\right)^n$ using the general binomial theorem. The convergence Condition becomes $\left|\frac{b}{a}x\right| \lt 1$.

Example

Expand $\sqrt{4 + x}$ in ascending powers of $x$ up to $x^2$. $$\begin{aligned} (4 + x)^{1/2} &= 4^{1/2}\left(1 + \frac{x}{4}\right)^{1/2} \\ &= 2\left[1 + \frac{1}{2}\cdot\frac{x}{4} + \frac{\frac{1}{2}\cdot\left(-\frac{1}{2}\right)}{2!}\cdot\frac{x^2}{16} + \cdots\right] \\ &= 2\left[1 + \frac{x}{8} - \frac{x^2}{128} + \cdots\right] \\ &= 2 + \frac{x}{4} - \frac{x^2}{64} + \cdots \end{aligned}$$

Valid for $\left|\frac{x}{4}\right| \lt 1$I.e., $|x| \lt 4$.

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In the expansion of $(a + bx)^n$The $(r+1)$Th term is:

$T_{r+1} = \binom{n}{r} a^{n-r}(bx)^r$

The index $r$ counts from $0$So the first term corresponds to $r = 0$ and the last term (when $n$ Is a positive integer) corresponds to $r = n$.

To find the coefficient of $x^k$ in $(a + bx)^n$:

1. Write the general term: $T_{r+1} = \binom{n}{r} a^{n-r} b^r x^r$.
2. Set the power of $x$ equal to $k$: $r = k$.
3. Substitute $r = k$ and evaluate.

When $n$ is not a positive integer, the general term still works — but you must also check Convergence ($|bx/a| \lt 1$).

Example 1: Constant term

Find the constant term in the expansion of $\left(x^2 + \frac{2}{x}\right)^8$.

The general term is:

$T_{r+1} = \binom{8}{r} (x^2)^{8-r}\left(\frac{2}{x}\right)^r = \binom{8}{r} 2^r x^{16-3r}$

For the constant term, set $16 - 3r = 0$Giving $r = \frac{16}{3}$. Since $r$ must be a Non-negative integer, there is no constant term.

(If the power had been $9$We would get $r = \frac{18}{3} = 6$And the constant term would be $\binom{9}{6} 2^6 = 84 \cdot 64 = 5376$.)

Example 2: Specific power of $x$

Find the coefficient of $x^5$ in the expansion of $(1 + 2x)^{12}$.

The general term is $T_{r+1} = \binom{12}{r}(2x)^r = \binom{12}{r} 2^r x^r$.

Set $r = 5$:

$\mathrm{Coefficient} = \binom{12}{5} 2^5 = 792 \cdot 32 = 25344$

Example 3: Non-integer power

Find the coefficient of $x^3$ in the expansion of $(1 - 3x)^{-1/2}$ up to $x^3$.

The general term is:

$T_{r+1} = \binom{-1/2}{r}(-3x)^r$

For $r = 3$:

$\binom{-1/2}{3} = \frac{(-1/2)(-3/2)(-5/2)}{3!} = \frac{-15/8}{6} = -\frac{5}{16}$

$T_4 = \left(-\frac{5}{16}\right)(-3x)^3 = \left(-\frac{5}{16}\right)(-27x^3) = \frac{135}{16}x^3$

Coefficient of $x^3$: $\frac{135}{16}$.

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When $(1 + x)^n$ is expanded as an infinite series, taking only the first $k + 1$ terms gives an Approximation. The accuracy depends on how many terms are taken and how small $|x|$ is.

For small $|x|$The series converges rapidly, so few terms are needed for high accuracy.

Theorem (Error bound for alternating decreasing series). If the terms of $(1+x)^n$ are Alternating in sign and decreasing in magnitude, then the error when truncating after $k$ terms is Bounded by the magnitude of the next (first omitted) term:

$\left|(1+x)^n - S_k\right| \leq |T_{k+1}|$

Where $S_k$ is the partial sum up to and including the $x^k$ term, and $T_{k+1}$ is the $(k+1)$Th Term.

This applies when $n \lt 0$ and $x \gt 0$Or when $n$ is fractional with alternating signs.

Example: Approximating $\sqrt[3]{28}$

We write $28 = 27 + 1 = 27(1 + 1/27)$So:

$\sqrt[3]{28} = \left(27\left(1 + \frac{1}{27}\right)\right)^{1/3} = 3\left(1 + \frac{1}{27}\right)^{1/3}$

Expanding with $n = 1/3$ and $x = 1/27$:

$$\begin{aligned} \left(1 + \frac{1}{27}\right)^{1/3} &= 1 + \frac{1}{3}\cdot\frac{1}{27} + \frac{\frac{1}{3}\cdot\left(-\frac{2}{3}\right)}{2!}\cdot\frac{1}{729} + \frac{\frac{1}{3}\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{5}{3}\right)}{3!}\cdot\frac{1}{19683} + \cdots \\ &= 1 + \frac{1}{81} - \frac{1}{6561} + \frac{5}{1594323} + \cdots \\ &\approx 1 + 0.012346 - 0.000152 + 0.000003 + \cdots \\ &\approx 1.012197 \end{aligned}$$

So $\sqrt[3]{28} \approx 3 \times 1.012197 = 3.036591$.

Error estimation. The terms alternate and decrease in magnitude. The next term (the $x^4$ term) Has magnitude:

$\left|\frac{\frac{1}{3}\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{5}{3}\right)\cdot\left(-\frac{8}{3}\right)}{4!}\cdot\frac{1}{27^4}\right| = \frac{80}{243 \cdot 531441} \approx 6.2 \times 10^{-7}$

So the error in the expansion of $(1 + 1/27)^{1/3}$ is at most $\approx 6.2 \times 10^{-7}$And the Error in $\sqrt[3]{28}$ is at most $3 \times 6.2 \times 10^{-7} \approx 1.9 \times 10^{-6}$.

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We already know (from Problem 7 below) that:

$\sum_{r=0}^{n} \binom{n}{r} = 2^n$

This follows from setting $a = 1$, $b = 1$ in the binomial theorem.

Theorem. $\displaystyle\sum_{r=0}^{n} r\binom{n}{r} = n \cdot 2^{n-1}$.

Proof (using differentiation).

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

Differentiate both sides with respect to $x$:

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

Multiply through by $x$:

$nx(1 + x)^{n-1} = \sum_{r=0}^{n} r\binom{n}{r} x^r$

Set $x = 1$:

$n \cdot 1 \cdot 2^{n-1} = \sum_{r=0}^{n} r\binom{n}{r}$

Therefore $\sum_{r=0}^{n} r\binom{n}{r} = n \cdot 2^{n-1}$. $\blacksquare$

Intuition. If you want to select a committee of any size from $n$ people and then choose a Chairperson, you can first pick $r$ members ($\binom{n}{r}$ ways) and then a chair from those $r$ ($r$ ways). Summing over all $r$ gives $\sum r\binom{n}{r}$. Alternatively, pick the chair first ($n$ ways) and then any subset of the remaining $n - 1$ people ($2^{n-1}$ ways), giving $n \cdot 2^{n-1}$.

Theorem. $\displaystyle\sum_{r=0}^{n} r^2\binom{n}{r} = n(n+1) \cdot 2^{n-2}$.

Proof (using differentiation).

We have $nx(1+x)^{n-1} = \sum_{r=0}^{n} r\binom{n}{r} x^r$.

Differentiate both sides with respect to $x$:

$n(1+x)^{n-1} + n(n-1)x(1+x)^{n-2} = \sum_{r=0}^{n} r^2\binom{n}{r} x^{r-1}$

Multiply through by $x$:

$nx(1+x)^{n-1} + n(n-1)x^2(1+x)^{n-2} = \sum_{r=0}^{n} r^2\binom{n}{r} x^r$

Set $x = 1$:

$n \cdot 2^{n-1} + n(n-1) \cdot 2^{n-2} = \sum_{r=0}^{n} r^2\binom{n}{r}$

Factor out $n \cdot 2^{n-2}$:

$n \cdot 2^{n-2}(2 + n - 1) = n(n+1) \cdot 2^{n-2} = \sum_{r=0}^{n} r^2\binom{n}{r} \quad \blacksquare$

Theorem (Vandermonde’s Identity). For non-negative integers $n, r, m$:

$\sum_{k=0}^{m} \binom{r}{k}\binom{n - r}{m - k} = \binom{n}{m}$

Proof (combinatorial).

Consider a set of $n$ people, of which $r$ are women and $n - r$ are men. The right-hand side $\binom{n}{m}$ counts the number of ways to choose a committee of $m$ people from all $n$.

Alternatively, we can count by cases: choose $k$ women and $m - k$ men. The number of ways to choose $k$ women is $\binom{r}{k}$And the number of ways to choose $m - k$ men is $\binom{n-r}{m-k}$. Summing over all valid $k$ gives the left-hand side. $\blacksquare$

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When expanding $(a + bx)^n$ with non-integer $n$We write it as $a^n\left(1 + \frac{bx}{a}\right)^n$. The expansion is valid when:

$\left|\frac{bx}{a}\right| \lt 1 \quad \mathrm{i.e.,} \quad |x| \lt \left|\frac{a}{b}\right|$

### 8.2 Product of Two Expansions

When a question asks for the expansion of a product of two binomial expressions, such as $(1 + px)^a(1 + qx)^b$Each factor has its own validity range:

$|px| \lt 1 \quad \mathrm{and} \quad |qx| \lt 1$

The combined expansion is valid only where both individual expansions are valid. This means the Overall validity is the intersection of the two ranges, which is always the more restrictive Condition:

$|x| \lt \min\left(\frac{1}{|p|}, \frac{1}{|q|}\right)$

Why? The product expansion is obtained by multiplying the individual series term by term. If Either series diverges, the term-by-term multiplication is not justified, so the product expansion May not equal the original expression.

For expressions involving three or more factors, the validity is the intersection of all individual Validity ranges.

For partial fraction decompositions that lead to sums of binomial expansions, the same principle Applies: the overall validity is the most restrictive individual condition.

Example 1

Find the expansion of $(1 + x)^{-2}(1 - 3x)^{-1}$ up to $x^2$Stating the validity.

$(1 + x)^{-2}$: valid for $|x| \lt 1$.

$(1 - 3x)^{-1}$: valid for $|3x| \lt 1$I.e., $|x| \lt \frac{1}{3}$.

The combined validity is $|x| \lt \frac{1}{3}$ (the more restrictive condition).

$(1 + x)^{-2} = 1 - 2x + 3x^2 + \cdots$

$(1 - 3x)^{-1} = 1 + 3x + 9x^2 + \cdots$

Product up to $x^2$:

$(1)(1) + [(-2)(1) + (1)(3)]x + [(3)(1) + (-2)(3) + (1)(9)]x^2$

$= 1 + x + (3 - 6 + 9)x^2 = 1 + x + 6x^2 + \cdots$

Example 2

Expand $\frac{3}{(1-x)(2+x)}$ in ascending powers of $x$ up to $x^2$.

Partial fractions: $\frac{3}{(1-x)(2+x)} = \frac{A}{1-x} + \frac{B}{2+x}$.

$3 = A(2 + x) + B(1 - x)$

Setting $x = 1$: $3 = 3A \implies A = 1$. Setting $x = -2$: $3 = 3B \implies B = 1$.

$\frac{3}{(1-x)(2+x)} = \frac{1}{1-x} + \frac{1}{2+x} = (1-x)^{-1} + \frac{1}{2}(1 + x/2)^{-1}$

$(1-x)^{-1}$: valid for $|x| \lt 1$.

$(1 + x/2)^{-1}$: valid for $|x/2| \lt 1$I.e., $|x| \lt 2$.

Combined validity: $|x| \lt 1$.

$(1-x)^{-1} = 1 + x + x^2 + \cdots$

$\frac{1}{2}\left(1 + \frac{x}{2}\right)^{-1} = \frac{1}{2}\left(1 - \frac{x}{2} + \frac{x^2}{4} - \cdots\right) = \frac{1}{2} - \frac{x}{4} + \frac{x^2}{8} - \cdots$

$\frac{3}{(1-x)(2+x)} = \frac{3}{2} + \frac{3}{4}x + \frac{9}{8}x^2 + \cdots$

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Problem 1. Expand $(1 + 3x)^4$.

Solution

$$(1 + 3x)^4 = \binom{4}{0} + \binom{4}{1}(3x) + \binom{4}{2}(3x)^2 + \binom{4}{3}(3x)^3 + \binom{4}{4}(3x)^4$$

$= 1 + 12x + 54x^2 + 108x^3 + 81x^4$

If you get this wrong, revise: [Binomial theorem](#3-the-binomial-theorem)

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

Solution

The $x^3$ term comes from $r = 3$:

$\binom{6}{3}(2)^{6-3}(-x)^3 = 20 \cdot 8 \cdot (-x^3) = -160x^3$

Coefficient of $x^3$: $-160$.

If you get this wrong, revise: [Binomial theorem](#3-the-binomial-theorem)

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Problem 3. Expand $(1 + x)^{-2}$ up to the term in $x^3$Stating the range of validity.

Solution

$$ \begin{aligned} (1 + x)^{-2} &= 1 + (-2)x + \frac{(-2)(-3)}{2!}x^2 + \frac{(-2)(-3)(-4)}{3!}x^3 + \cdots \\ &= 1 - 2x + 3x^2 - 4x^3 + \cdots \end{aligned} $$

Valid for $|x| \lt 1$.

If you get this wrong, revise: [General binomial expansion](#4-general-binomial-expansion)

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Problem 4. Find the coefficient of $x^2$ in the expansion of $(1 - 2x)^5(1 + 3x)^4$.

Solution

$(1 - 2x)^5 = 1 - 10x + 40x^2 + \cdots$

$(1 + 3x)^4 = 1 + 12x + 54x^2 + \cdots$

Coefficient of $x^2$ in the product:

• From $(1)(54x^2)$: $54$
• From $(-10x)(12x)$: $-120$
• From $(40x^2)(1)$: $40$

Total: $54 - 120 + 40 = -26$.

If you get this wrong, revise: [Binomial theorem](#3-the-binomial-theorem)

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Problem 5. Use the binomial expansion to estimate $\sqrt{1.05}$ to 5 decimal places.

Solution

$$\sqrt{1.05} = (1 + 0.05)^{1/2}$$

$= 1 + \frac{1}{2}(0.05) + \frac{\frac{1}{2} \cdot \left(-\frac{1}{2}\right)}{2}(0.05)^2 + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}(0.05)^3 + \cdots$

$= 1 + 0.025 - 0.0003125 + 0.0000078125 - \cdots$

$\approx 1.024695$

To 5 d.p.: $1.02470$.

If you get this wrong, revise: [General binomial expansion](#4-general-binomial-expansion)

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Problem 6. In the expansion of $(1 + ax)^n$The coefficients of $x$, $x^2$And $x^3$ are in The ratio $1 : 4 : 12$. Find $a$ and $n$.

Solution

Coefficient of $x$: $na$

Coefficient of $x^2$: $\frac{n(n-1)}{2}a^2$

Coefficient of $x^3$: $\frac{n(n-1)(n-2)}{6}a^3$

Ratio $1 : 4 : 12$:

$\frac{n(n-1)}{2}a^2 = 4na \implies \frac{(n-1)a}{2} = 4 \implies (n-1)a = 8 \quad \mathrm{--- (1)}$

$\frac{n(n-1)(n-2)}{6}a^3 = 12na \implies \frac{(n-1)(n-2)a^2}{6} = 12 \quad \mathrm{--- (2)}$

From (1): $a = \frac{8}{n-1}$. Substitute into (2):

$\frac{(n-1)(n-2)}{6} \cdot \frac{64}{(n-1)^2} = 12$

$\frac{64(n-2)}{6(n-1)} = 12$

$64(n-2) = 72(n-1)$

$64n - 128 = 72n - 72$

$-8n = 56 \implies n = -7$

$a = \frac{8}{-8} = -1$.

If you get this wrong, revise: [General binomial expansion](#4-general-binomial-expansion)

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Problem 7. Prove that $\sum_{r=0}^{n} \binom{n}{r} = 2^n$.

Solution

Setting $a = 1$ and $b = 1$ in the binomial theorem:

$(1 + 1)^n = \sum_{r=0}^{n} \binom{n}{r} 1^{n-r} 1^r = \sum_{r=0}^{n} \binom{n}{r}$

So $\sum_{r=0}^{n} \binom{n}{r} = 2^n$. $\blacksquare$

Intuition. The sum of all binomial coefficients counts the total number of subsets of an $n$-element set, which is $2^n$ (each element can either be included or excluded).

If you get this wrong, revise: [Binomial coefficients](#2-binomial-coefficients)

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Problem 8. Expand $\frac{1}{(1 + x)(1 - 2x)}$ in ascending powers of $x$ up to $x^3$Stating The range of validity.

Solution

Using partial fractions: $\frac{1}{(1+x)(1-2x)} = \frac{1}{3}\cdot\frac{1}{1+x} + \frac{2}{3}\cdot\frac{1}{1-2x}$.

$\frac{1}{1+x} = (1+x)^{-1} = 1 - x + x^2 - x^3 + \cdots \quad (|x| \lt 1)$

$\frac{1}{1-2x} = (1-2x)^{-1} = 1 + 2x + 4x^2 + 8x^3 + \cdots \quad (|2x| \lt 1)$

$$\begin{aligned} \frac{1}{(1+x)(1-2x)} &= \frac{1}{3}(1 - x + x^2 - x^3) + \frac{2}{3}(1 + 2x + 4x^2 + 8x^3) + \cdots \\ &= \frac{1}{3} + \frac{2}{3} + \left(-\frac{1}{3} + \frac{4}{3}\right)x + \left(\frac{1}{3} + \frac{8}{3}\right)x^2 + \left(-\frac{1}{3} + \frac{16}{3}\right)x^3 + \cdots \\ &= 1 + x + 3x^2 + 5x^3 + \cdots \end{aligned}$$

Valid for $|x| \lt \frac{1}{2}$ (the more restrictive condition).

If you get this wrong, revise: [General binomial expansion](#4-general-binomial-expansion)

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Problem 9. Given that the expansion of $(1 + kx)^{10}$ in ascending powers of $x$ has a Coefficient of $x^3$ equal to $1080$Find the value of $k$.

Solution

$$\binom{10}{3} k^3 = 1080$$

$120k^3 = 1080$

$k^3 = 9$

$k = \sqrt[3]{9}$

If you get this wrong, revise: [Binomial theorem](#3-the-binomial-theorem)

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Problem 10. Prove that $\sum_{r=0}^{n} (-1)^r \binom{n}{r} = 0$ for $n \geq 1$.

Solution

Setting $a = 1$ and $b = -1$ in the binomial theorem:

$(1 - 1)^n = \sum_{r=0}^{n} \binom{n}{r} 1^{n-r}(-1)^r = \sum_{r=0}^{n} (-1)^r \binom{n}{r}$

$0^n = 0$ for $n \geq 1$. $\blacksquare$

Intuition. This counts the difference between subsets of even size and subsets of odd size — which Is zero by a parity argument (there’s a bijection between even-sized and odd-sized subsets: add or Remove one element).

If you get this wrong, revise: [Binomial coefficients](#2-binomial-coefficients)

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Problem 11. Find the coefficient of $x^3$ in the expansion of $(1 + x)^{-3}(1 - 2x)^{-1}$ up to $x^3$Stating the range of validity.

Solution

First expand each factor:

$(1 + x)^{-3} = 1 + (-3)x + \frac{(-3)(-4)}{2}x^2 + \frac{(-3)(-4)(-5)}{6}x^3 + \cdots = 1 - 3x + 6x^2 - 10x^3 + \cdots$

$(1 - 2x)^{-1} = 1 + 2x + 4x^2 + 8x^3 + \cdots$

Coefficient of $x^3$ in the product:

• From $(1)(8x^3)$: $8$
• From $(-3x)(4x^2)$: $-12$
• From $(6x^2)(2x)$: $12$
• From $(-10x^3)(1)$: $-10$

Total: $8 - 12 + 12 - 10 = -2$.

Validity: $(1+x)^{-3}$ requires $|x| \lt 1$ and $(1-2x)^{-1}$ requires $|x| \lt 1/2$. Overall: $|x| \lt 1/2$.

If you get this wrong, revise: [Validity of general binomial expansion](#8-validity-of-general-binomial-expansion)

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Problem 12. Use the binomial expansion to approximate $\sqrt[3]{126}$ to 4 decimal places. Estimate the error in your approximation.

Solution

Write $126 = 125 + 1 = 125(1 + 1/125)$:

$\sqrt[3]{126} = 5\left(1 + \frac{1}{125}\right)^{1/3}$

Expand with $n = 1/3$, $x = 1/125 = 0.008$:

$$\begin{aligned} \left(1 + \frac{1}{125}\right)^{1/3} &= 1 + \frac{1}{3}\cdot\frac{1}{125} + \frac{\frac{1}{3}\cdot\left(-\frac{2}{3}\right)}{2}\cdot\frac{1}{15625} + \cdots \\ &= 1 + \frac{1}{375} - \frac{1}{70312.5} + \cdots \\ &\approx 1 + 0.0026667 - 0.0000142 + \cdots \\ &\approx 1.0026525 \end{aligned}$$

$\sqrt[3]{126} \approx 5 \times 1.0026525 = 5.01326$

To 4 d.p.: $5.0133$.

Error estimate. The terms alternate and decrease. The next term (the $x^3$ term) has magnitude:

$\left|\frac{\frac{1}{3}\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{5}{3}\right)}{6}\cdot\frac{1}{125^3}\right| = \frac{10/162}{6} \cdot \frac{1}{1953125} \approx 5.3 \times 10^{-9}$

The error in $\sqrt[3]{126}$ is at most $5 \times 5.3 \times 10^{-9} \approx 2.7 \times 10^{-8}$ Which is negligible for 4 decimal places. The approximation $5.0133$ is reliable.

If you get this wrong, revise: [Binomial approximation and error estimation](#6-binomial-approximation-and-error-estimation)

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Problem 13. Evaluate $\sum_{r=0}^{10} r\binom{10}{r}$.

Solution

By the identity $\sum_{r=0}^{n} r\binom{n}{r} = n \cdot 2^{n-1}$ with $n = 10$:

$\sum_{r=0}^{10} r\binom{10}{r} = 10 \cdot 2^{9} = 10 \times 512 = 5120$

Verification by differentiation:

$(1+x)^{10} = \sum_{r=0}^{10} \binom{10}{r} x^r$. Differentiate and set $x = 1$:

$10(1+1)^9 = \sum_{r=0}^{10} r\binom{10}{r} = 10 \times 512 = 5120$ ✓

If you get this wrong, revise: [Relationships between binomial coefficients](#7-relationships-between-binomial-coefficients)

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Problem 14. In the expansion of $(2 + 3x)^{2n}$The ratio of the coefficient of $x^3$ to the Coefficient of $x$ is $11 : 1$. Find the value of $n$.

Solution

The general term in the expansion of $(2 + 3x)^{2n}$ is $\binom{2n}{r} 2^{2n-r}(3x)^r$.

Coefficient of $x$: $\binom{2n}{1} 2^{2n-1} \cdot 3 = 2n \cdot 2^{2n-1} \cdot 3 = 3n \cdot 2^{2n}$.

Coefficient of $x^3$: $\binom{2n}{3} 2^{2n-3} \cdot 27 = \frac{2n(2n-1)(2n-2)}{6} \cdot 27 \cdot 2^{2n-3}$.

Ratio is $11 : 1$:

$\frac{\displaystyle \frac{2n(2n-1)(2n-2)}{6} \cdot 27 \cdot 2^{2n-3}}{3n \cdot 2^{2n}} = 11$

Simplify:

$\frac{2n(2n-1)(2n-2) \cdot 27 \cdot 2^{2n-3}}{6 \cdot 3n \cdot 2^{2n}} = 11$

$\frac{2n(2n-1)(2n-2) \cdot 9 \cdot 2^{2n-3}}{6n \cdot 2^{2n}} = 11$

$\frac{(2n-1)(2n-2) \cdot 18 \cdot 2^{2n-3}}{6 \cdot 2^{2n}} = 11$

$\frac{(2n-1)(2n-2) \cdot 18}{6 \cdot 8} = 11$

$\frac{(2n-1)(2n-2) \cdot 3}{8} = 11$

$(2n-1)(2n-2) = \frac{88}{3}$

This gives a non-integer, so let us recheck. We need $n$ such that the ratio equals $11$. Trying Small values:

$n = 3$: coefficient of $x^3$ is $\binom{6}{3} 2^3 \cdot 27 = 20 \cdot 8 \cdot 27 = 4320$. Coefficient of $x$ is $6 \cdot 32 \cdot 3 = 576$. Ratio: $4320/576 = 7.5$.

$n = 4$: coefficient of $x^3$ is $\binom{8}{3} 2^5 \cdot 27 = 56 \cdot 32 \cdot 27 = 48384$. Coefficient of $x$ is $8 \cdot 128 \cdot 3 = 3072$. Ratio: $48384/3072 \approx 15.75$.

$n = 5$: coefficient of $x^3$ is $\binom{10}{3} 2^7 \cdot 27 = 120 \cdot 128 \cdot 27 = 414720$. Coefficient of $x$ is $10 \cdot 512 \cdot 3 = 15360$. Ratio: $414720/15360 = 27$.

Since $7.5 \lt 11 \lt 15.75$ and the ratio is increasing, there is no integer $n$ giving ratio Exactly $11$. However, solving the equation more carefully:

$\frac{(2n-1)(2n-2)}{4} = 11$

$2(2n-1)(n-1) = 11$

$(2n-1)(n-1) = \frac{11}{2}$

$2n^2 - 3n + 1 = \frac{11}{2}$

$4n^2 - 6n - 9 = 0$

$n = \frac{6 \pm \sqrt{36 + 144}}{8} = \frac{6 \pm \sqrt{180}}{8} = \frac{6 \pm 6\sqrt{5}}{8} = \frac{3 \pm 3\sqrt{5}}{4}$

Since $n$ must be a positive integer, there is no integer solution. The ratio $11 : 1$ cannot be Achieved for any positive integer $n$ with $(2+3x)^{2n}$.

(The closest integer value is $n = 4$ giving ratio $\approx 15.75$And $n = 3$ giving $7.5$.)

If you get this wrong, revise: [Finding specific terms](#5-finding-specific-terms)

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Problem 15. Expand $\frac{2}{(1+x)^2(1-2x)}$ in ascending powers of $x$ up to $x^2$Stating the Range of validity.

Solution

First, find partial fractions. Write:

$\frac{2}{(1+x)^2(1-2x)} = \frac{A}{1+x} + \frac{B}{(1+x)^2} + \frac{C}{1-2x}$

$2 = A(1+x)(1-2x) + B(1-2x) + C(1+x)^2$

Setting $x = -2$: $2 = C(-1)^2 = C$So $C = 2$.

Setting $x = -1$: $2 = B(3)$So $B = \frac{2}{3}$.

Setting $x = 0$: $2 = A + B + C = A + 2/3 + 2$So $A = 2 - 8/3 = -2/3$.

So:

$\frac{2}{(1+x)^2(1-2x)} = -\frac{2}{3}(1+x)^{-1} + \frac{2}{3}(1+x)^{-2} + 2(1-2x)^{-1}$

Now expand each:

$(1+x)^{-1} = 1 - x + x^2 + \cdots \quad (|x| \lt 1)$

$(1+x)^{-2} = 1 - 2x + 3x^2 + \cdots \quad (|x| \lt 1)$

$(1-2x)^{-1} = 1 + 2x + 4x^2 + \cdots \quad (|x| \lt 1/2)$

$$\begin{aligned} \frac{2}{(1+x)^2(1-2x)} &= -\frac{2}{3}(1 - x + x^2) + \frac{2}{3}(1 - 2x + 3x^2) + 2(1 + 2x + 4x^2) + \cdots \\ &= \left(-\frac{2}{3} + \frac{2}{3} + 2\right) + \left(\frac{2}{3} - \frac{4}{3} + 4\right)x + \left(-\frac{2}{3} + 2 + 8\right)x^2 + \cdots \\ &= 2 + \frac{10}{3}x + \frac{28}{3}x^2 + \cdots \end{aligned}$$

Validity: the most restrictive condition is $|x| \lt 1/2$ (from $(1-2x)^{-1}$).

If you get this wrong, revise: [Validity of general binomial expansion](#8-validity-of-general-binomial-expansion)

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• Sequences and Series — Binomial expansions for negative or fractional powers produce infinite series, directly linking to convergence of series.
• Proof — The binomial theorem is proved by induction, and binomial coefficient identities appear frequently in proof questions.
• Differentiation — The derivative of $x^n$ is proved using the binomial theorem applied to $(x+h)^n$.
• Integration — Binomial expansions are used to integrate functions that cannot be handled by standard rules.