Sequences and Series

Board Coverage

Board	Paper	Notes
AQA	Paper 1, 2	Arithmetic and geometric sequences, sigma notation
Edexcel	P1, P2	Same; recurrence relations in P2
OCR (A)	Paper 1	Arithmetic and geometric progressions
CIE (9709)	P1, P3	Sequences and series; P3 includes $\Sigma$ notation more extensively

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1. Sequences and Series: Definitions

Definition. A sequence is an ordered list of numbers $(a_1, a_2, a_3, \ldots)$. We write $(a_n)_{n=1}^{\infty}$ or simply $(a_n)$.

Definition. A series is the sum of the terms of a sequence: $\sum_{n=1}^{N} a_n = a_1 + a_2 + \cdots + a_N$.

Definition. A sequence defined by $a_{n+1} = f(a_n)$ with an initial value $a_1$ is a recurrence relation (or iterative sequence).

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2. Arithmetic Sequences

Definition. An arithmetic sequence (arithmetic progression, AP) is a sequence where each term differs from the previous by a constant $d$ (the common difference):

$a_{n+1} = a_n + d$

2.1 The $n$th Term

Theorem. The $n$th term of an arithmetic sequence with first term $a$ and common difference $d$ is:

$a_n = a + (n - 1)d$

Proof. By induction on $n$.

Base case ($n = 1$): $a_1 = a + 0 \cdot d = a$. ✓

Inductive step: Assume $a_k = a + (k - 1)d$. Then:

$a_{k+1} = a_k + d = a + (k - 1)d + d = a + kd$

This matches the formula for $n = k + 1$. $\blacksquare$

2.2 Sum of an Arithmetic Series

Theorem. The sum of the first $n$ terms of an arithmetic sequence is:

$S_n = \frac{n}{2}(2a + (n - 1)d) = \frac{n}{2}(a + \ell)$

where $\ell = a_n = a + (n - 1)d$ is the last term.

Proof (Pairing Method). Write out the sum twice, once forwards and once backwards:

$$\begin{aligned} S_n &= a + (a + d) + (a + 2d) + \cdots + (a + (n-1)d) \\ S_n &= (a + (n-1)d) + (a + (n-2)d) + \cdots + a \end{aligned}$$

Adding vertically, each pair sums to $2a + (n-1)d$, and there are $n$ such pairs:

$2S_n = n(2a + (n - 1)d)$

$S_n = \frac{n}{2}(2a + (n - 1)d) \quad \blacksquare$

Intuition. Gauss supposedly used this method as a child to sum $1 + 2 + \cdots + 100 = 5050$. Pair the first and last, second and second-to-last, etc. Each pair sums to the same value.

Details

Example

Find the sum of the first 20 terms of $3, 7, 11, 15, \ldots$

Here $a = 3$, $d = 4$, $n = 20$.

$S_{20} = \frac{20}{2}(2 \times 3 + 19 \times 4) = 10(6 + 76) = 10 \times 82 = 820$

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3. Geometric Sequences

Definition. A geometric sequence (geometric progression, GP) is a sequence where each term is a constant multiple $r$ (the common ratio) of the previous term:

$a_{n+1} = a_n \cdot r$

3.1 The $n$th Term

Theorem. The $n$th term of a geometric sequence with first term $a$ and common ratio $r$ is:

$a_n = ar^{n-1}$

Proof. By induction.

Base case: $a_1 = ar^0 = a$. ✓

Inductive step: $a_{k+1} = a_k \cdot r = ar^{k-1} \cdot r = ar^k$. ✓ $\blacksquare$

3.2 Sum of a Finite Geometric Series

Theorem. For $r \neq 1$:

$S_n = a\frac{1 - r^n}{1 - r} = a\frac{r^n - 1}{r - 1}$

Proof. Write:

$$\begin{aligned} S_n &= a + ar + ar^2 + \cdots + ar^{n-1} \\ rS_n &= ar + ar^2 + ar^3 + \cdots + ar^n \end{aligned}$$

Subtracting: $S_n - rS_n = a - ar^n$

$S_n(1 - r) = a(1 - r^n)$

$S_n = \frac{a(1 - r^n)}{1 - r} \quad \blacksquare$

Intuition (Self-Similarity). Multiplying the sum by $r$ shifts every term one position to the right. The original sum and the shifted sum overlap almost completely — the difference is just the first term minus the new last term. This "shift and subtract" idea is the same principle behind many iterative algorithms.

3.3 Sum to Infinity

Theorem. If $|r| < 1$, the infinite geometric series converges, and:

$S_\infty = \sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1 - r}$

Proof. From $S_n = \frac{a(1 - r^n)}{1 - r}$, we take the limit as $n \to \infty$.

Since $|r| < 1$, we have $\lim_{n \to \infty} r^n = 0$ (a standard limit; see below).

$S_\infty = \lim_{n \to \infty} S_n = \frac{a(1 - 0)}{1 - r} = \frac{a}{1 - r} \quad \blacksquare$

Lemma. If $|r| < 1$, then $\lim_{n \to \infty} r^n = 0$.

Proof. Write $r^n = e^{n \ln|r|}$. Since $|r| < 1$, we have $\ln|r| < 0$. As $n \to \infty$, $n \ln|r| \to -\infty$, so $e^{n \ln|r|} \to 0$. $\blacksquare$

Theorem. If $|r| \geq 1$, the geometric series $\sum_{n=1}^{\infty} ar^{n-1}$ diverges.

Proof. If $|r| > 1$, then $|r^n| \to \infty$, so $|a_n| \to \infty$. Since the terms don't tend to zero, the series diverges by the divergence test.

If $r = 1$: $S_n = na \to \pm\infty$ (unless $a = 0$).

If $r = -1$: $S_n = a - a + a - a + \cdots$, which oscillates and does not converge. $\blacksquare$

warning

The condition $|r| < 1$ is essential. A common mistake is to apply the sum-to-infinity formula when $|r| \geq 1$, which gives nonsense.

Details

Example

Find the sum to infinity of $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$

Here $a = 1$, $r = \frac{1}{2}$, $|r| < 1$.

$S_\infty = \frac◆LB◆1◆RB◆◆LB◆1 - \frac{1}{2}◆RB◆ = 2$

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4. Sigma Notation

Definition. $\sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n$.

Properties:

$$\begin{aligned} \sum_{k=1}^{n} (a_k + b_k) &= \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k \\ \sum_{k=1}^{n} ca_k &= c\sum_{k=1}^{n} a_k \end{aligned}$$

4.1 Standard Sums

$$\begin{aligned} \sum_{k=1}^{n} k &= \frac{n(n+1)}{2} \\ \sum_{k=1}^{n} k^2 &= \frac{n(n+1)(2n+1)}{6} \\ \sum_{k=1}^{n} k^3 &= \frac{n^2(n+1)^2}{4} \end{aligned}$$

Proof of $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$. This is the arithmetic series with $a = 1$, $d = 1$, $n$ terms. By the formula: $S_n = \frac{n}{2}(2 + (n-1)) = \frac{n(n+1)}{2}$. $\blacksquare$

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5. Recurrence Relations

A recurrence relation defines each term in terms of previous terms. A recurrence relation of order $k$ requires $k$ initial conditions.

Example. $u_{n+1} = 2u_n + 3$, $u_1 = 1$.

$u_2 = 2(1) + 3 = 5$, $u_3 = 2(5) + 3 = 13$, $u_4 = 2(13) + 3 = 29$, ...

Periodic sequences. If $u_{n+1} = f(u_n)$ and the sequence returns to a previous value, it becomes periodic.

Example. $u_{n+1} = \frac{1}{u_n}$, $u_1 = 2$.

$u_2 = \frac{1}{2}$, $u_3 = 2$, $u_4 = \frac{1}{2}$, ... This is periodic with period 2.

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6. Arithmetic Mean and Geometric Mean

Definition. The arithmetic mean (AM) of two positive numbers $a$ and $b$ is $\frac{a+b}{2}$. The geometric mean (GM) is $\sqrt{ab}$.

Theorem (AM-GM Inequality). For any positive real numbers $a, b$:

$\frac{a + b}{2} \geq \sqrt{ab}$

Equality holds if and only if $a = b$.

Proof. Since $a, b \gt{} 0$, both $\sqrt{a}$ and $\sqrt{b}$ are real numbers. For any real number $x$, we have $x^2 \geq 0$. In particular:

$(\sqrt{a} - \sqrt{b})^2 \geq 0$

Expanding:

$a - 2\sqrt{a}\sqrt{b} + b \geq 0$

$a + b \geq 2\sqrt{ab}$

$\frac{a + b}{2} \geq \sqrt{ab} \quad \blacksquare$

Equality condition. $(\sqrt{a} - \sqrt{b})^2 = 0$ if and only if $\sqrt{a} = \sqrt{b}$, i.e., $a = b$.

Extension. For $n$ positive real numbers $x_1, x_2, \ldots, x_n$:

$\frac◆LB◆x_1 + x_2 + \cdots + x_n◆RB◆◆LB◆n◆RB◆ \geq \sqrt[n]{x_1 x_2 \cdots x_n}$

with equality if and only if $x_1 = x_2 = \cdots = x_n$. The proof of the general case (by induction using the two-variable result as the base) is beyond A-level scope.

Details

Example

Find the minimum value of $x + \frac{4}{x}$ for $x \gt{} 0$.

By AM-GM with $a = x$ and $b = \frac{4}{x}$ (both positive):

$\frac◆LB◆x + \frac{4}{x}◆RB◆◆LB◆2◆RB◆ \geq \sqrt◆LB◆x \cdot \frac{4}{x}◆RB◆ = \sqrt{4} = 2$

So $x + \frac{4}{x} \geq 4$.

Equality when $x = \frac{4}{x}$, i.e., $x^2 = 4$, so $x = 2$ (since $x \gt{} 0$).

Minimum value is 4, achieved at $x = 2$.

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7. Sigma Notation — Method of Differences

Definition. A telescoping sum is a series where most terms cancel when written out, leaving only a few terms at the beginning and end.

Key Idea. If we can express the general term $u_k$ as a difference $f(k) - f(k+1)$, then:

$\sum_{k=1}^{n} u_k = \sum_{k=1}^{n} [f(k) - f(k+1)] = f(1) - f(n+1)$

This is because the sum expands as $[f(1) - f(2)] + [f(2) - f(3)] + \cdots + [f(n) - f(n+1)]$, and all intermediate terms cancel.

7.1 Partial Fractions as the Decomposition Tool

The most common technique is to use partial fractions to decompose a rational term into a difference.

Details

Example

Evaluate $\sum_{k=1}^{n} \frac{1}{k(k+1)}$.

Using partial fractions:

$\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}$

$1 = A(k+1) + Bk$

Setting $k = 0$: $A = 1$. Setting $k = -1$: $B = -1$.

So $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.

Therefore:

$\sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+1}\right)$

$= \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$

$= 1 - \frac{1}{n+1} = \frac{n}{n+1}$

Details

Example

Evaluate $\sum_{k=1}^{n} \frac{1}{k(k+2)}$.

Using partial fractions:

$\frac{1}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2}$

$1 = A(k+2) + Bk$

$k = 0$: $A = \frac{1}{2}$. $k = -2$: $B = -\frac{1}{2}$.

So $\frac{1}{k(k+2)} = \frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+2}\right)$.

$\sum_{k=1}^{n} \frac{1}{k(k+2)} = \frac{1}{2}\sum_{k=1}^{n}\left(\frac{1}{k} - \frac{1}{k+2}\right)$

Writing out terms:

$= \frac{1}{2}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+2}\right)\right]$

The terms $-\frac{1}{3}$ and $+\frac{1}{3}$ cancel, $-\frac{1}{4}$ and $+\frac{1}{4}$ cancel, etc. The surviving terms are $1$ and $\frac{1}{2}$ from the start, with $-\frac{1}{n+1}$ and $-\frac{1}{n+2}$ at the end:

$= \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right) = \frac{1}{2}\left(\frac{3}{2} - \frac{2n + 3}{(n+1)(n+2)}\right)$

$= \frac{3}{4} - \frac{2n + 3}{2(n+1)(n+2)}$

tip

When using the method of differences, always write out the first few terms explicitly to identify the cancellation pattern before simplifying. Be especially careful when the "gap" in the denominator is larger than 1 (e.g., $k(k+2)$), as not all terms cancel in a simple pairwise fashion.

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8. Arithmetic-Geometric Sequences

Definition. An arithmetic-geometric sequence has terms of the form:

$a, \; (a+d)r, \; (a+2d)r^2, \; \ldots$

Each term is the product of a term from an arithmetic progression ($a, a+d, a+2d, \ldots$) and a term from a geometric progression ($1, r, r^2, \ldots$).

Theorem. The $n$th term is:

$u_n = (a + (n-1)d)\,r^{n-1}$

where $a$ is the first term of the AP part, $d$ is the common difference, and $r$ is the common ratio of the GP part.

8.1 Sum of an Arithmetic-Geometric Series

Theorem. For $r \neq 1$:

$S_n = \frac{a - [a + (n-1)d]\,r^n}{1 - r} + \frac{dr(1 - r^{n-1})}{(1-r)^2}$

Proof. Write out $S_n$ and $rS_n$, then subtract:

$$\begin{aligned} S_n &= a + (a+d)r + (a+2d)r^2 + \cdots + [a+(n-1)d]\,r^{n-1} \\ rS_n &= ar + (a+d)r^2 + (a+2d)r^3 + \cdots + [a+(n-1)d]\,r^n \end{aligned}$$

Subtracting:

$S_n - rS_n = a + dr + dr^2 + \cdots + dr^{n-1} - [a+(n-1)d]\,r^n$

$S_n(1-r) = a + d(r + r^2 + \cdots + r^{n-1}) - [a+(n-1)d]\,r^n$

The bracketed geometric series sums to $\frac{r(1 - r^{n-1})}{1 - r}$:

$S_n(1-r) = a + \frac{dr(1 - r^{n-1})}{1-r} - [a+(n-1)d]\,r^n$

$S_n = \frac{a - [a+(n-1)d]\,r^n}{1-r} + \frac{dr(1 - r^{n-1})}{(1-r)^2} \quad \blacksquare$

8.2 Sum to Infinity

When $|r| \lt{} 1$, both $r^n \to 0$ and $r^{n-1} \to 0$ as $n \to \infty$:

$S_\infty = \frac{a}{1 - r} + \frac{dr}{(1-r)^2}$

Details

Example

A salary is 30000 in year 1 and increases by 1500 each year. Due to inflation, each year's salary is discounted by a factor of 0.9 when expressed in present-value terms. Find the total present value of all future salary payments.

The sequence of discounted salaries is an arithmetic-geometric sequence:

• AP part: $a = 30000$, $d = 1500$
• GP part: $r = 0.9$

Since $|r| \lt{} 1$:

$S_\infty = \frac{30000}{1 - 0.9} + \frac◆LB◆1500 \times 0.9◆RB◆◆LB◆(1 - 0.9)^2◆RB◆$

$= \frac{30000}{0.1} + \frac{1350}{0.01}$

$= 300000 + 135000 = 435000$

The total present value is 435000.

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9. Proof of Sum Formulas

9.1 Proof of $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

Proof by induction.

Base case ($n = 1$): LHS $= 1$. RHS $= \frac◆LB◆1 \times 2 \times 3◆RB◆◆LB◆6◆RB◆ = 1$. ✓

Inductive step: Assume $\displaystyle\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ for some $n \geq 1$.

Then:

$$\begin{aligned} \sum_{k=1}^{n+1} k^2 &= \frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\ &= \frac{n(n+1)(2n+1) + 6(n+1)^2}{6} \\ &= \frac◆LB◆(n+1)\bigl[n(2n+1) + 6(n+1)\bigr]◆RB◆◆LB◆6◆RB◆ \\ &= \frac{(n+1)(2n^2 + 7n + 6)}{6} \\ &= \frac{(n+1)(n+2)(2n+3)}{6} \end{aligned}$$

This equals $\frac◆LB◆(n+1)\bigl((n+1)+1\bigr)\bigl(2(n+1)+1\bigr)◆RB◆◆LB◆6◆RB◆$, which is the formula for $n+1$. ✓ $\blacksquare$

9.2 Proof of $\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$

Proof by induction.

Base case ($n = 1$): LHS $= 1$. RHS $= \left[\frac◆LB◆1 \times 2◆RB◆◆LB◆2◆RB◆\right]^2 = 1$. ✓

Inductive step: Assume $\displaystyle\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$.

Then:

$$\begin{aligned} \sum_{k=1}^{n+1} k^3 &= \left[\frac{n(n+1)}{2}\right]^2 + (n+1)^3 \\ &= \frac{n^2(n+1)^2}{4} + (n+1)^3 \\ &= \frac{n^2(n+1)^2 + 4(n+1)^3}{4} \\ &= \frac◆LB◆(n+1)^2\bigl(n^2 + 4(n+1)\bigr)◆RB◆◆LB◆4◆RB◆ \\ &= \frac{(n+1)^2(n^2 + 4n + 4)}{4} \\ &= \frac{(n+1)^2(n+2)^2}{4} \\ &= \left[\frac{(n+1)(n+2)}{2}\right]^2 \end{aligned}$$

This is the formula for $n+1$. ✓ $\blacksquare$

9.3 Connection: $(\sum k)^2 = \sum k^3$

Notice that:

$\left(\sum_{k=1}^{n} k\right)^2 = \left[\frac{n(n+1)}{2}\right]^2 = \sum_{k=1}^{n} k^3$

This is a remarkable identity: the square of the sum of the first $n$ positive integers equals the sum of the first $n$ cubes.

Pattern observation. Check for small values of $n$:

$n = 1$: $(1)^2 = 1 = 1^3$ ✓

$n = 2$: $(1+2)^2 = 9 = 1 + 8 = 1^3 + 2^3$ ✓

$n = 3$: $(1+2+3)^2 = 36 = 1 + 8 + 27 = 1^3 + 2^3 + 3^3$ ✓

This can also be visualised geometrically: a square of side $\frac{n(n+1)}{2}$ can be decomposed into nested gnomons (L-shaped regions) that correspond to $1^3, 2^3, \ldots, n^3$.

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10. Problem Set

Problem 1. The 5th term of an arithmetic sequence is 17 and the 12th term is 38. Find the first term and the common difference.

Details

Solution

$a + 4d = 17 \quad \mathrm{--- (1)}$ $a + 11d = 38 \quad \mathrm{--- (2)}$

(2) - (1): $7d = 21 \implies d = 3$.

$a = 17 - 12 = 5$.

If you get this wrong, revise: Arithmetic sequences

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Problem 2. Evaluate $\sum_{k=1}^{50} (3k - 1)$.

Details

Solution

This is an arithmetic series with first term $a = 2$, last term $\ell = 3(50) - 1 = 149$, $n = 50$.

$S = \frac{50}{2}(2 + 149) = 25 \times 151 = 3775$

If you get this wrong, revise: Sigma notation

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Problem 3. A geometric series has first term 5 and sum to infinity 25. Find the common ratio.

Details

Solution

$S_\infty = \frac{a}{1 - r} = 25$

$\frac{5}{1 - r} = 25 \implies 1 - r = \frac{1}{5} \implies r = \frac{4}{5}$

If you get this wrong, revise: Sum to infinity

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Problem 4. Find the sum of the first 10 terms of the geometric series $2 - 6 + 18 - 54 + \cdots$

Details

Solution

$a = 2$, $r = -3$, $n = 10$.

$S_{10} = \frac{2(1 - (-3)^{10})}{1 - (-3)} = \frac{2(1 - 59049)}{4} = \frac{2(-59048)}{4} = -29524$

If you get this wrong, revise: Sum of finite geometric series

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Problem 5. Show that $\sum_{k=1}^{n} (4k + 1) = n(2n + 3)$.

Solution

$$\begin{aligned} \sum_{k=1}^{n} (4k + 1) &= 4\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \\ &= 4 \cdot \frac{n(n+1)}{2} + n \\ &= 2n(n+1) + n \\ &= 2n^2 + 2n + n \\ &= 2n^2 + 3n \\ &= n(2n + 3) \quad \blacksquare \end{aligned}$$

If you get this wrong, revise: Sigma notation

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Problem 6. Given $u_1 = 3$ and $u_{n+1} = \frac{u_n + 1}{u_n - 1}$, find $u_2$, $u_3$, $u_4$, and $u_5$. Comment on the sequence.

Details

Solution

$u_2 = \frac{3 + 1}{3 - 1} = 2$

$u_3 = \frac{2 + 1}{2 - 1} = 3$

$u_4 = \frac{3 + 1}{3 - 1} = 2$

$u_5 = \frac{2 + 1}{2 - 1} = 3$

The sequence is periodic: $3, 2, 3, 2, 3, 2, \ldots$ with period 2.

If you get this wrong, revise: Recurrence relations

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Problem 7. The first three terms of a geometric sequence are $x, x + 4, x + 12$. Find $x$ and the common ratio.

Details

Solution

$\frac{x + 4}{x} = \frac{x + 12}{x + 4}$

$(x + 4)^2 = x(x + 12)$

$x^2 + 8x + 16 = x^2 + 12x$

$4x = 16 \implies x = 4$

The sequence is $4, 8, 16, \ldots$ with $r = 2$.

If you get this wrong, revise: Geometric sequences

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Problem 8. A ball is dropped from a height of 10 m. Each bounce reaches 80% of the previous height. Find the total distance travelled before the ball comes to rest.

Details

Solution

The ball falls 10 m, then rises $10 \times 0.8 = 8$ m, falls 8 m, rises $8 \times 0.8 = 6.4$ m, etc.

Total distance = $10 + 2(8 + 6.4 + 5.12 + \cdots)$.

The bracketed series is geometric with $a = 8$, $r = 0.8$.

$S_\infty = \frac{8}{1 - 0.8} = \frac{8}{0.2} = 40$

Total distance = $10 + 2 \times 40 = 90$ m.

If you get this wrong, revise: Sum to infinity

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Problem 9. Find the least value of $n$ such that the sum of the first $n$ terms of $3 + 6 + 12 + 24 + \cdots$ exceeds 10000.

Details

Solution

$a = 3$, $r = 2$.

$S_n = \frac{3(2^n - 1)}{2 - 1} = 3(2^n - 1) > 10000$

$2^n - 1 > \frac{10000}{3} \implies 2^n > \frac{10003}{3} \approx 3334.33$

$n > \log_2(3334.33) \approx 11.7$

So $n = 12$.

Check: $S_{11} = 3(2048 - 1) = 6141 < 10000$.

$S_{12} = 3(4096 - 1) = 12285 > 10000$. ✓

If you get this wrong, revise: Sum of finite geometric series

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Problem 10. Prove that $\sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(n+2)}{3}$.

Details

Solution

By induction.

Base case ($n = 1$): LHS $= 1 \times 2 = 2$. RHS $= \frac◆LB◆1 \times 2 \times 3◆RB◆◆LB◆3◆RB◆ = 2$. ✓

Inductive step: Assume $\sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(n+2)}{3}$.

Then:

$$\begin{aligned} \sum_{k=1}^{n+1} k(k+1) &= \frac{n(n+1)(n+2)}{3} + (n+1)(n+2) \\ &= \frac{n(n+1)(n+2) + 3(n+1)(n+2)}{3} \\ &= \frac{(n+1)(n+2)(n + 3)}{3} \\ &= \frac{(n+1)((n+1)+1)((n+1)+2)}{3} \end{aligned}$$

This is the formula for $n + 1$. ✓ $\blacksquare$

If you get this wrong, revise: Proof by induction

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Problem 11. Given that $x \gt{} 0$, find the minimum value of $x^2 + \frac{9}{x^2}$ and state the value of $x$ at which it occurs.

Details

Solution

By AM-GM with $a = x^2$ and $b = \frac{9}{x^2}$ (both positive since $x \gt{} 0$):

$\frac◆LB◆x^2 + \frac{9}{x^2}◆RB◆◆LB◆2◆RB◆ \geq \sqrt◆LB◆x^2 \cdot \frac{9}{x^2}◆RB◆ = \sqrt{9} = 3$

So $x^2 + \frac{9}{x^2} \geq 6$.

Equality when $x^2 = \frac{9}{x^2}$, i.e., $x^4 = 9$, so $x^2 = 3$, giving $x = \sqrt{3}$ (positive root).

Minimum value is 6, achieved at $x = \sqrt{3}$.

If you get this wrong, revise: AM-GM inequality

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Problem 12. Evaluate $\sum_{k=1}^{n} \frac{2}{(k+1)(k+3)}$ using the method of differences.

Details

Solution

Using partial fractions:

$\frac{2}{(k+1)(k+3)} = \frac{A}{k+1} + \frac{B}{k+3}$

$2 = A(k+3) + B(k+1)$

$k = -3$: $B = -1$. $k = -1$: $A = 1$.

So $\frac{2}{(k+1)(k+3)} = \frac{1}{k+1} - \frac{1}{k+3}$.

$\sum_{k=1}^{n} \frac{2}{(k+1)(k+3)} = \sum_{k=1}^{n} \left(\frac{1}{k+1} - \frac{1}{k+3}\right)$

Writing out terms:

$= \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{4} - \frac{1}{6}\right) + \cdots + \left(\frac{1}{n+1} - \frac{1}{n+3}\right)$

After cancellation, surviving terms:

$= \frac{1}{2} + \frac{1}{3} - \frac{1}{n+2} - \frac{1}{n+3} = \frac{5}{6} - \frac{2n + 5}{(n+2)(n+3)}$

If you get this wrong, revise: Method of differences

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Problem 13. Find the sum to infinity of the arithmetic-geometric series whose terms are $1, \; 4 \times \tfrac{1}{2}, \; 7 \times \tfrac{1}{4}, \; 10 \times \tfrac{1}{8}, \; \ldots$

Details

Solution

Identify the components:

• AP part: first term $a = 1$, common difference $d = 3$ (since $4 - 1 = 3$, $7 - 4 = 3$, etc.)
• GP part: common ratio $r = \frac{1}{2}$

Since $|r| \lt{} 1$, the sum to infinity converges:

$S_\infty = \frac{a}{1 - r} + \frac{dr}{(1-r)^2}$

$= \frac◆LB◆1◆RB◆◆LB◆1 - \frac{1}{2}◆RB◆ + \frac◆LB◆3 \times \frac{1}{2}◆RB◆◆LB◆\left(1 - \frac{1}{2}\right)^2◆RB◆$

$= \frac◆LB◆1◆RB◆◆LB◆\frac{1}{2}◆RB◆ + \frac◆LB◆\frac{3}{2}◆RB◆◆LB◆\frac{1}{4}◆RB◆$

$= 2 + 6 = 8$

If you get this wrong, revise: Arithmetic-geometric sequences

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Problem 14. Find $\sum_{k=1}^{n} k(k-1)$ in closed form, and verify your answer for $n = 4$.

Solution

$$\begin{aligned} \sum_{k=1}^{n} k(k-1) &= \sum_{k=1}^{n} (k^2 - k) \\ &= \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k \\ &= \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} \\ &= \frac{n(n+1)}{6}\bigl[(2n+1) - 3\bigr] \\ &= \frac{n(n+1)(2n - 2)}{6} \\ &= \frac{n(n+1)(n-1)}{3} \end{aligned}$$

Verification for $n = 4$: $1 \times 0 + 2 \times 1 + 3 \times 2 + 4 \times 3 = 0 + 2 + 6 + 12 = 20$.

Formula: $\frac◆LB◆4 \times 5 \times 3◆RB◆◆LB◆3◆RB◆ = 20$. ✓

If you get this wrong, revise: Sigma notation

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Problem 15. A sequence satisfies $u_{n+1} = 3u_n + 2$ with $u_1 = 1$. Find a closed-form expression for $u_n$ and verify it for $n = 1, 2, 3$.

Details

Solution

This is a first-order linear recurrence relation. We solve it by finding the equilibrium and subtracting.

At equilibrium, $u = 3u + 2$, giving $-2u = 2$, so $u = -1$.

Define $v_n = u_n - (-1) = u_n + 1$. Then:

$v_{n+1} = u_{n+1} + 1 = 3u_n + 2 + 1 = 3u_n + 3 = 3(u_n + 1) = 3v_n$

So $v_n$ is a geometric sequence with ratio 3. Since $v_1 = u_1 + 1 = 2$:

$v_n = 2 \cdot 3^{n-1}$

Therefore:

$u_n = 2 \cdot 3^{n-1} - 1$

Verification:

• $n = 1$: $u_1 = 2 \cdot 1 - 1 = 1$ ✓
• $n = 2$: $u_2 = 2 \cdot 3 - 1 = 5$. Check: $3(1) + 2 = 5$ ✓
• $n = 3$: $u_3 = 2 \cdot 9 - 1 = 17$. Check: $3(5) + 2 = 17$ ✓

If you get this wrong, revise: Recurrence relations

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tip

tip Ready to test your understanding of Sequences and Series? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Sequences and Series with other pure mathematics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.