4.3 Sequences and Functions

$$a_n = a + (n-1)d$$ where:
• $a_n$ = the nth term
• $a$ = first term
• $n$ = position in sequence
• $d$ = common difference

Step 1: Find the common difference
$d = 7 - 3 = 4$

Step 2: Identify the first term
$a = 3$

Step 3: Substitute into the formula
$a_n = 3 + (n-1)(4)$
$a_n = 3 + 4n - 4$
$a_n = 4n - 1$

Check: When $n=1$: $4(1) - 1 = 3$ ✓

$a = 5$, $d = 3$, $n = 50$

$a_{50} = 5 + (50-1)(3)$
$a_{50} = 5 + 49 \times 3$
$a_{50} = 5 + 147 = 152$

Step 1: Find $d$ and write the formula
$d = 7$, so $a_n = 2 + (n-1)(7) = 7n - 5$

Step 2: Set equal to 142 and solve
$7n - 5 = 142$
$7n = 147$
$n = 21$

Answer: 142 is the 21st term

$$a_n = a \times r^{n-1}$$ where:
• $a_n$ = the nth term
• $a$ = first term
• $r$ = common ratio
• $n$ = position in sequence

$a = 2$, $r = 3$, $n = 6$

$a_6 = 2 \times 3^{6-1}$
$a_6 = 2 \times 3^5$
$a_6 = 2 \times 243 = 486$

Step 1: Find $r$
$r = \frac{20}{100} = 0.2$ (or $\frac{1}{5}$)

Step 2: Find the 5th term
$a_5 = 100 \times (0.2)^4$
$a_5 = 100 \times 0.0016 = 0.16$

Arithmetic: +/- the same number each time → linear growth

Geometric: ×/÷ the same number each time → exponential growth/decay

Step 1: Find the second difference → this equals $2a$
Step 2: Write $an^2$ and subtract from the sequence
Step 3: Find the nth term of what remains (it's arithmetic!)
Step 4: Combine to get $an^2 + bn + c$

Step 1: Find differences
1st diff: 5, 7, 9, 11 (not constant)
2nd diff: 2, 2, 2 (constant!) → $2a = 2$, so $a = 1$

Step 2: Subtract $n^2$ from the sequence

$n$	1	2	3	4	5
Sequence	3	8	15	24	35
$n^2$	1	4	9	16	25
Difference	2	4	6	8	10

Step 3: The remaining sequence (2, 4, 6, 8, 10) is $2n$

Answer: $a_n = n^2 + 2n$
Check: $n=3$: $9 + 6 = 15$ ✓

Step 1: 2nd differences = 2, so $a = 1$

Step 2: Subtract $n^2$:
Sequence: 0, 3, 8, 15, 24
$n^2$: 1, 4, 9, 16, 25
Difference: -1, -1, -1, -1, -1

Step 3: Remaining is just $-1$

Answer: $a_n = n^2 - 1$
Notice: This is $(n+1)(n-1)$!

$f(x) = 2x + 3$ means "the function $f$ takes $x$ and returns $2x + 3$"

$f(5)$ means "substitute $x = 5$" → $f(5) = 2(5) + 3 = 13$

a) $f(3)$:
$f(3) = (3)^2 - 4(3) + 1 = 9 - 12 + 1 = -2$

b) $f(-2)$:
$f(-2) = (-2)^2 - 4(-2) + 1 = 4 + 8 + 1 = 13$

c) $f(0)$:
$f(0) = 0 - 0 + 1 = 1$

$3x - 1 = 14$
$3x = 15$
$x = 5$

$fg(x)$ means: do $g$ first, then $f$

Read right to left! $fg(x) = f(g(x))$

Step 1: Do $g$ first
$g(3) = 3^2 = 9$

Step 2: Apply $f$ to the result
$f(9) = 2(9) + 1 = 19$

Answer: $fg(3) = 19$

Step 1: Do $f$ first
$f(3) = 2(3) + 1 = 7$

Step 2: Apply $g$ to the result
$g(7) = 7^2 = 49$

Answer: $gf(3) = 49$

Notice: $fg(3) \neq gf(3)$ - order matters!

Step 1: Replace $x$ in $f$ with $g(x)$
$fg(x) = f(g(x)) = f(x + 4)$

Step 2: Substitute into $f$
$= 3(x + 4) - 2$
$= 3x + 12 - 2$
$= 3x + 10$

If $f(a) = b$, then $f^{-1}(b) = a$

$f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$

Step 1: Write $y = f(x)$
Step 2: Swap $x$ and $y$
Step 3: Rearrange to make $y$ the subject
Step 4: Replace $y$ with $f^{-1}(x)$

Step 1: $y = 2x + 5$

Step 2: Swap: $x = 2y + 5$

Step 3: Rearrange for $y$
$x - 5 = 2y$
$y = \frac{x - 5}{2}$

Answer: $f^{-1}(x) = \frac{x - 5}{2}$

Step 1: $y = \frac{3x - 1}{4}$

Step 2: Swap: $x = \frac{3y - 1}{4}$

Step 3: Rearrange
$4x = 3y - 1$
$4x + 1 = 3y$
$y = \frac{4x + 1}{3}$

Answer: $g^{-1}(x) = \frac{4x + 1}{3}$

Step 1: $y = x^2 + 3$

Step 2: Swap: $x = y^2 + 3$

Step 3: Rearrange
$x - 3 = y^2$
$y = \sqrt{x - 3}$ (taking positive root since $x \geq 0$)

Answer: $h^{-1}(x) = \sqrt{x - 3}$

• Any integer: $n$
• Even number: $2n$
• Odd number: $2n + 1$
• Consecutive integers: $n, n+1, n+2, ...$
• Consecutive even: $2n, 2n+2, 2n+4, ...$
• Consecutive odd: $2n+1, 2n+3, 2n+5, ...$
• Multiple of 3: $3n$

Step 1: Let the two odd numbers be $2a + 1$ and $2b + 1$

Step 2: Add them
$(2a + 1) + (2b + 1) = 2a + 2b + 2$

Step 3: Factorise
$= 2(a + b + 1)$

Step 4: Conclude
This is 2 times an integer, so it's even. ∎

Step 1: Let the integers be $n$, $n+1$, $n+2$

Step 2: Add them
$n + (n+1) + (n+2) = 3n + 3$

Step 3: Factorise
$= 3(n + 1)$

Step 4: Conclude
This is 3 times an integer, so it's divisible by 3. ∎

Method 1: Factorising
$n^2 + n = n(n + 1)$
This is the product of two consecutive integers.
One of them must be even, so the product is even. ∎

Method 2: Cases
If $n$ is even ($n = 2k$): $n^2 + n = 4k^2 + 2k = 2(2k^2 + k)$ → even ✓
If $n$ is odd ($n = 2k+1$): $n^2 + n = (2k+1)^2 + (2k+1) = 4k^2 + 4k + 2 = 2(2k^2 + 2k + 1)$ → even ✓

LHS:
$(n+1)^2 - (n-1)^2$
$= (n^2 + 2n + 1) - (n^2 - 2n + 1)$
$= n^2 + 2n + 1 - n^2 + 2n - 1$
$= 4n$

LHS = RHS, so the statement is proven. ∎

Alternative: Use DOTS! $(n+1)^2 - (n-1)^2 = [(n+1)+(n-1)][(n+1)-(n-1)] = 2n \times 2 = 4n$