Sequences and Series

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

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

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

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 $n = k + 1$. $\blacksquare$

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.

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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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:

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

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

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.

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. $\blacksquare$

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.

$S_\infty = \frac{1}{1 - \frac{1}{2}} = 2$

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

Properties:

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

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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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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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 $x_1, x_2, \ldots, x_n$:

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

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.

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):

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

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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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.

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

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:

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

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

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

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:

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

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

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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.

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 $\frac{r(1 - r^{n-1})}{1 - r}$:

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

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

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

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

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 $|r| \lt 1$:

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

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

$= 300000 + 135000 = 435000$

The total present value is 435000.

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Proof by induction.

Base case ($n = 1$): LHS $= 1$. RHS $= \frac{1 \times 2 \times 3}{6} = 1$. ✓

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

Then:

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

This equals $\frac{(n+1)\bigl((n+1)+1\bigr)\bigl(2(n+1)+1\bigr)}{6}$Which is the formula for $n+1$. ✓ $\blacksquare$

Proof by induction.

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

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

Then:

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

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

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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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.

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](#2-arithmetic-sequences)

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

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](#4-sigma-notation)

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

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](#33-sum-to-infinity)

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

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](#32-sum-of-a-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](#4-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.

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](#5-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.

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](#3-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.

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](#33-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.

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](#32-sum-of-a-finite-geometric-series)

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

Solution

By induction.

Base case ($n = 1$): LHS $= 1 \times 2 = 2$. RHS $= \frac{1 \times 2 \times 3}{3} = 2$. ✓

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](./13-proof.md)

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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.

Solution

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

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

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](#6-arithmetic-mean-and-geometric-mean)

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

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](#7-sigma-notation--method-of-differences)

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

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:

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

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

$= \frac{1}{\frac{1}{2}} + \frac{\frac{3}{2}}{\frac{1}{4}}$

$= 2 + 6 = 8$

If you get this wrong, revise: [Arithmetic-geometric sequences](#8-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: $\frac{4 \times 5 \times 3}{3} = 20$. ✓

If you get this wrong, revise: [Sigma notation](#4-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 $n = 1, 2, 3$.

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](#5-recurrence-relations)

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