Good morning fellow learners! Ever set up a line of dominoes? You tip the first one, and it sets off a chain reaction, each domino knocking over the next in a predictable pattern. In the world of mathematics, we have something very similar: recursive formulas. Instead of giving you a map to jump straight to the 100th domino, a recursive formula gives you the simple, powerful instruction for how any one domino knocks over the next. It’s all about understanding the step-by-step relationship that brings a sequence to life.
In this guide, we’re going to demystify these powerful tools. We’ll start with the basics, move through common patterns you’ll see in your IB course (both AI and AA!), and even level up to some trickier problems that require a bit of algebraic detective work. Ready to start the chain reaction? Let’s go!
Table of Contents
What Are Recursive Formulas, Anyway?
Think of it like a treasure hunt. An explicit formula is like a GPS that tells you the exact coordinates of the treasure ($u_{100}$). A recursive formula is more like a set of clues: “From wherever you are ($u_{n-1}$), take three steps east to find the next spot ($u_n$).” You can’t find the 100th spot without visiting the 99th first. This step-by-step nature is the heart of recursion.
The Two Golden Rules
Every complete recursive definition for a sequence must have two parts. No exceptions!
- The Recursion Rule: This is the formula that connects a term to its predecessor(s). It’s the “how-to” part, usually written in terms of $u_n$ and $u_{n-1}$ (or even $u_{n-2}$).
- The Initial Condition(s): This is the starting point. It’s the first term of the sequence, written as $u_1 = [some value]$. Without it, your rule has no place to begin. It’s like having instructions for the treasure hunt but no starting address!
Cracking the Code: Finding the Formula
Finding the recursive formula is all about pattern recognition. Let’s look at the most common types you’ll encounter.
Case 1: The Steady Adder (Arithmetic Sequences)
This is the most straightforward case. If you’re adding (or subtracting) the same number every time to get to the next term, you’ve got an arithmetic sequence.
Example: Arithmetic Sequence
Find the recursive formula for the sequence: $10, 6, 2, -2, \dots$
Step 1: Find the pattern. Let’s check the difference between consecutive terms.
- $6 – 10 = -4$
- $2 – 6 = -4$
- $-2 – 2 = -4$
Aha! We’re consistently subtracting 4. This is our common difference.
Step 2: Write the rule. To get any term ($u_n$), we take the previous term ($u_{n-1}$) and subtract 4. So, the rule is $u_n = u_{n-1} – 4$.
Step 3: State the initial condition. The sequence starts with 10, so $u_1 = 10$.
Complete Formula: $$u_n = u_{n-1} – 4, \quad u_1 = 10$$
Case 2: The Constant Multiplier (Geometric Sequences)
If you’re multiplying (or dividing) by the same number each time, you’re dealing with a geometric sequence.
Example: Geometric Sequence
Find the recursive formula for the sequence: $80, 40, 20, 10, \dots$
Step 1: Find the pattern. Adding won’t work here. Let’s check the ratio between consecutive terms.
- $40 / 80 = 0.5$
- $20 / 40 = 0.5$
- $10 / 20 = 0.5$
Looks like we’re consistently multiplying by 0.5. This is our common ratio.
Step 2: Write the rule. To get any term ($u_n$), we take the previous term ($u_{n-1}$) and multiply by 0.5. The rule is $u_n = 0.5 u_{n-1}$.
Step 3: State the initial condition. The sequence starts at 80, so $u_1 = 80$.
Complete Formula: $$u_n = 0.5 u_{n-1}, \quad u_1 = 80$$
Case 3: The Pattern-Within-a-Pattern
Sometimes the difference isn’t constant, but the differences themselves form a pattern! This often points to a quadratic-style sequence. The trick is to look at the “second differences.”
Example: Quadratic-like Sequence
Find the recursive formula for the sequence: $2, 5, 10, 17, 26, \dots$
Step 1: Find the first differences.
Sequence: 2 5 10 17 26
Differences: +3 +5 +7 +9
The first differences ($3, 5, 7, 9, \dots$) aren’t constant. But wait… they form their own arithmetic sequence!
Step 2: Find the second differences. Let’s find the differences of our new sequence.
First Differences: 3 5 7 9
Second Differences: +2 +2 +2
A constant second difference! This is our key. It tells us the amount we add each time is increasing linearly.
Step 3: Write the rule. We know $u_n = u_{n-1} + (\text{something})$. That “something” is the $(n-1)$-th term of our first difference sequence. The first differences are $3, 5, 7, \dots$, which is an arithmetic sequence with first term 3 and common difference 2. Its general term is $d_k = 3 + (k-1)2 = 2k+1$. The amount we add to get from $u_{n-1}$ to $u_n$ is the $(n-1)$-th difference, so we use $d_{n-1} = 2(n-1)+1 = 2n-1$.
So, our rule is $u_n = u_{n-1} + (2n-1)$. Let’s test it: to get $u_4=17$, we take $u_3=10$ and add $2(4)-1=7$. $10+7=17$. It works!
Step 4: State the initial condition. The sequence begins with 2, so $u_1=2$.
Complete Formula: $$u_n = u_{n-1} + 2n – 1, \quad \text{for } n \ge 2, \quad u_1 = 2$$
Leveling Up: Second-Order Recurrence Relations
What happens when a term depends not just on the one before it, but on the two before it? Welcome to second-order recurrence relations! These often appear in the form $u_{n+2} = a u_{n+1} + b u_n$, where our job is to find the constants ‘a’ and ‘b’.
The Game Plan: Solving for Two Unknowns
This might look scary, but it’s a classic IB problem-solving setup. The secret weapon? Simultaneous linear equations.
- Use the first three terms ($u_1, u_2, u_3$) to create your first equation by setting $n=1$ in the general form.
- Use the next set of terms ($u_2, u_3, u_4$) to create your second equation by setting $n=2$.
- Solve the system of two equations for your two unknowns, ‘a’ and ‘b’.
Example: Second-Order Relation
A sequence has terms $1, 3, 11, 43, \dots$ and follows the rule $u_{n+2} = a u_{n+1} + b u_n$. Find the values of $a$ and $b$.
We are given $u_1=1, u_2=3, u_3=11, u_4=43$.
Step 1: Form Equation 1 (using $n=1$).
The formula for $n=1$ is $u_3 = a u_2 + b u_1$. Plugging in our values:
$$11 = a(3) + b(1) \implies 3a + b = 11 \quad (Eq. 1)$$
Step 2: Form Equation 2 (using $n=2$).
The formula for $n=2$ is $u_4 = a u_3 + b u_2$. Plugging in our values:
$$43 = a(11) + b(3) \implies 11a + 3b = 43 \quad (Eq. 2)$$
Step 3: Solve the system. We can use substitution. From Eq. 1, it’s easy to isolate $b$: $b = 11 – 3a$.
Now, substitute this expression for $b$ into Eq. 2:
$$11a + 3(11 – 3a) = 43$$
$$11a + 33 – 9a = 43$$
$$2a = 10$$
$$a = 5$$
Finally, find $b$ by substituting $a=5$ back into our expression for $b$: $b = 11 – 3(5) = 11 – 15 = -4$.
Complete Formula: We found $a=5$ and $b=-4$. The recursive formula is $$u_{n+2} = 5u_{n+1} – 4u_n$$ and you would also need to state the initial conditions $u_1=1, u_2=3$.
Key Takeaways
Phew, that was a lot, but you’ve got this! Here are the core ideas to lock in:
- Recursive formulas define a term based on the ones that came before it. It’s a step-by-step process.
- Never, ever forget your initial condition(s)! A rule without a starting point is incomplete.
- Be a pattern detective! Look for a common difference (arithmetic), a common ratio (geometric), or patterns in the differences themselves.
- For second-order relations ($u_{n+2} = a u_{n+1} + b u_n$), your go-to strategy is always to build and solve a system of simultaneous equations.
Have questions or want to discuss a problem? Share your thoughts in the comments below! Engaging with the material and your peers is a fantastic way to deepen your understanding and analytical skills in mathematics.
Additional Practice
Question 1: Find the recursive formula for the sequence $100, 93, 86, 79, \dots$
Show Answer
Solution: This is an arithmetic sequence with a common difference of $93-100=-7$. The first term is 100.
Formula: $u_n = u_{n-1} – 7$, with $u_1 = 100$.
Question 2: A sequence is defined by $u_n = u_{n-1} + 12$ and $u_1 = -5$. Find the first four terms.
Show Answer
Solution:
$u_1 = -5$
$u_2 = u_1 + 12 = -5 + 12 = 7$
$u_3 = u_2 + 12 = 7 + 12 = 19$
$u_4 = u_3 + 12 = 19 + 12 = 31$
The first four terms are -5, 7, 19, 31.
Question 3: Find the recursive formula for the sequence $3, 4.5, 6, 7.5, \dots$
Show Answer
Solution: This is an arithmetic sequence with a common difference of $4.5 – 3 = 1.5$. The first term is 3.
Formula: $u_n = u_{n-1} + 1.5$, with $u_1 = 3$.
Question 4: Find the recursive formula for the sequence $5, 15, 45, 135, \dots$
Show Answer
Solution: This is a geometric sequence with a common ratio of $15/5=3$. The first term is 5.
Formula: $u_n = 3u_{n-1}$, with $u_1 = 5$.
Question 5: Find the recursive formula for the sequence $128, 64, 32, 16, \dots$
Show Answer
Solution: This is a geometric sequence with a common ratio of $64/128=0.5$. The first term is 128.
Formula: $u_n = 0.5u_{n-1}$, with $u_1 = 128$.
Question 6: A sequence is defined by $u_n = (-2)u_{n-1}$ and $u_1 = 7$. Find the fourth term, $u_4$.
Show Answer
Solution:
$u_1 = 7$
$u_2 = (-2)u_1 = -2(7) = -14$
$u_3 = (-2)u_2 = -2(-14) = 28$
$u_4 = (-2)u_3 = -2(28) = -56$
The fourth term is -56.
Question 7: Find the recursive formula for the sequence $0, 4, 10, 18, 28, \dots$
Show Answer
Solution:
First differences: $4, 6, 8, 10, \dots$
Second differences: $2, 2, 2, \dots$
The first differences form an arithmetic sequence $d_k = 4+(k-1)2 = 2k+2$. The amount to add to get from $u_{n-1}$ to $u_n$ is the $(n-1)$-th term of this difference sequence, which is $d_{n-1} = 2(n-1)+2 = 2n$.
Formula: $u_n = u_{n-1} + 2n$ for $n \ge 2$, with $u_1 = 0$.
Question 8: Find the recursive formula for the sequence $1, 2, 5, 10, 17, \dots$
Show Answer
Solution:
First differences: $1, 3, 5, 7, \dots$
Second differences: $2, 2, 2, \dots$
The first differences form an arithmetic sequence $d_k = 1+(k-1)2 = 2k-1$. The amount to add to get from $u_{n-1}$ to $u_n$ is the $(n-1)$-th term of this difference sequence, which is $d_{n-1} = 2(n-1)-1 = 2n-3$.
Formula: $u_n = u_{n-1} + 2n – 3$ for $n \ge 2$, with $u_1 = 1$.
Question 9: A sequence has terms $2, 5, 17, 65, \dots$ and follows the rule $u_{n+2} = a u_{n+1} + b u_n$. Find the values of $a$ and $b$.
Show Answer
Solution: We have $u_1=2, u_2=5, u_3=17, u_4=65$.
For $n=1$: $u_3 = au_2 + bu_1 \implies 17 = 5a + 2b$ (Eq. 1)
For $n=2$: $u_4 = au_3 + bu_2 \implies 65 = 17a + 5b$ (Eq. 2)
Multiply Eq. 1 by 5 and Eq. 2 by 2 to eliminate $b$:
$85 = 25a + 10b$
$130 = 34a + 10b$
Subtracting the first from the second: $45 = 9a \implies a=5$.
Substitute $a=5$ into Eq. 1: $17 = 5(5) + 2b \implies 17 = 25 + 2b \implies -8 = 2b \implies b=-4$.
So, $a=5$ and $b=-4$.
IB Exam Style Question
A sequence is defined recursively by $u_{n+2} = p u_{n+1} + q u_n$. Given $u_1 = 4, u_2 = 2, u_3 = 8,$ and $u_4 = -4$.
a) Find the values of $p$ and $q$.
b) Hence, find the value of $u_5$.
Reveal Solution
Solution:
a) Finding p and q
We need to set up a system of two equations using the given terms.
For $n=1$: $u_3 = p u_2 + q u_1 \implies 8 = p(2) + q(4) \implies 2p + 4q = 8$. Dividing by 2 simplifies this to $p + 2q = 4$ (Eq. 1).
For $n=2$: $u_4 = p u_3 + q u_2 \implies -4 = p(8) + q(2) \implies 8p + 2q = -4$. Dividing by 2 simplifies this to $4p + q = -2$ (Eq. 2).
Now we solve the system. From Eq. 1, we can write $p = 4 – 2q$. Substitute this into Eq. 2:
$$4(4 – 2q) + q = -2$$
$$16 – 8q + q = -2$$
$$16 – 7q = -2$$
$$-7q = -18 \implies q = 18/7$$
Now substitute $q$ back into the expression for $p$:
$$p = 4 – 2(18/7) = 28/7 – 36/7 = -8/7$$
So, $p = -8/7$ and $q = 18/7$.
b) Finding u_5
We use the formula with $n=3$: $u_5 = p u_4 + q u_3$.
$$u_5 = (-8/7)(-4) + (18/7)(8)$$
$$u_5 = 32/7 + 144/7$$
$$u_5 = 176/7$$