Answer key · Unit 13

Sequences and series.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 20 minutes

Arithmetic sequences

A sequence that adds the same number at every step: the general term, how to find the difference, how Gauss added the numbers from 1 to 100 in pairs, and why the odd numbers always add up to a square.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Add the difference
Practice · Predict, then look
Question
$a_1 = 1$ ⁠ and $d = 2$ ⁠. What are the first five terms of the sequence?

Correct answer
- Correct $1, 3, 5, 7, 9$ ⁠
Setup, choices and feedback
The Scene at this step
Difference d
2fixed
Ratio q
0.5defaulthidden
Terms n
1
Add the terms
Termsfixed
Choices
- Correct $1, 3, 5, 7, 9$ ⁠
- Wrong $2, 4, 6, 8, 10$ ⁠
  Feedback
  That sequence starts at $2$ ⁠. Here the first term is $a_1 = 1$ ⁠, and the $2$ ⁠ is added only from the second term on.
- Wrong $1, 2, 4, 8, 16$ ⁠
  Feedback
  Here you multiplied by $2$ ⁠. In an arithmetic sequence you add the difference $d$ ⁠ at every step.
- Wrong $1, 2, 3, 4, 5$ ⁠
  Feedback
  That is the sequence with $d = 1$ ⁠. Here every step adds $2$ ⁠.
Once the prediction is locked in, the Scene changes
Terms n
5
The Scene then shows: 9 · The last arithmetic term shown, aₙ = a₁ + (n − 1)d/nodes/stairs/outputs/arithmeticLast
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$1, 3, 5, 7, 9$ ⁠: start at $1$ ⁠ and add $2$ ⁠ at every step. These are the odd numbers, and the columns climb in a straight line.

Worked solution
1. The first term is given: $a_1 = 1$ ⁠.
2. Each term is $d = 2$ ⁠ more than the one before: $a_2 = 3$ ⁠, $a_3 = 5$ ⁠, $a_4 = 7$ ⁠, $a_5 = 9$ ⁠.
If you chose $1, 2, 4, 8, 16$ ⁠, you multiplied instead of adding: that is a geometric sequence.
Step 2 of 5
Open this step
The general term
Practice · Reach the target
Question
$a_1 = 1$ ⁠ and $n = 8$ ⁠ are locked. Change Difference d so that the eighth term is $a_8 = 22$ ⁠, then press Check.

Goal
- 22 ± 0.01
  At the start of the step: 8
Setup, choices and feedback
The Scene at this step
Difference d
1
Ratio q
0.5defaulthidden
Terms n
8fixed
Add the terms
Termsfixed
Read from the Scene at
The last arithmetic term shown, aₙ = a₁ + (n − 1)d/nodes/stairs/outputs/arithmeticLast
One setting that reaches it
Difference d
3
Feedback for a right answer
$a_8 = 1 + 7d = 22$ ⁠, so $7d = 21$ ⁠ and $d = 3$ ⁠. From the first term to the eighth there are seven steps, not eight.

Feedback for a wrong answer
Not yet. How many steps are there from $a_1$ ⁠ to $a_8$ ⁠, and how much must each one climb?

Hint
$a_8 = a_1 + 7d$ ⁠.

Worked solution
1. The general term: $a_n = a_1 + (n - 1)d$ ⁠.
2. Put in $n = 8$ ⁠: $a_8 = 1 + 7d$ ⁠.
3. $1 + 7d = 22$ ⁠, so $7d = 21$ ⁠ and $d = 3$ ⁠.
4. Set Difference d to $3$ ⁠.
If you tried $d = \frac{21}{8}$ ⁠, you counted eight steps: the first term is zero steps along.
Step 3 of 5
Open this step
Terms far away
Practice · Calculate
Question
In the sequence of odd numbers, $a_1 = 1$ ⁠ and $d = 2$ ⁠. The board shows only ten terms. What is the $100$ ⁠th term?

Correct answer
199 ± 0
Setup, choices and feedback
The Scene at this step
Difference d
2fixed
Ratio q
0.5defaulthidden
Terms n
10
Add the terms
Terms
Accepted within
± 0
A fixed value in the question
Wrong answers it recognises
- 201 ± 0
  Misconception: Counts n steps to the nth term instead of n − 1, so the general term comes out one step too far.
  Feedback from the catalogue
  From the first term to the $n$ ⁠th there are only $n - 1$ ⁠ steps: the first term itself is zero steps along. So $a_n = a_1 + (n - 1)d$ ⁠ for an arithmetic sequence and $a_n = a_1 q^{n-1}$ ⁠ for a geometric one. Check with $n = 1$ ⁠: the formula must give $a_1$ ⁠.
- 200 ± 0
  Misconception: Counts n steps to the nth term instead of n − 1, so the general term comes out one step too far.
  Feedback
  $100 \cdot 2 = 200$ ⁠ assumes the sequence starts at $0$ ⁠. It starts at $a_1 = 1$ ⁠, and $2$ ⁠ is added to it $99$ ⁠ times.
Feedback for a right answer
$a_{100} = 1 + 99 \cdot 2 = 199$ ⁠. Without the formula you would have to add $2$ ⁠ ninety-nine times.

Hint
How many steps of $2$ ⁠ are there from $a_1$ ⁠ to $a_{100}$ ⁠?

Worked solution
1. $a_n = a_1 + (n - 1)d$ ⁠.
2. $a_{100} = 1 + (100 - 1) \cdot 2 = 1 + 198 = 199$ ⁠.
If you got $201$ ⁠, you added $d$ ⁠ a hundred times: from the first term to the hundredth there are only $99$ ⁠ steps. If you got $200$ ⁠, you forgot that the sequence starts at $1$ ⁠.
Step 4 of 5
Open this step
Adding in pairs
Practice · Calculate
Question
$a_1 = 1$ ⁠, $d = 1$ ⁠ and $n = 10$ ⁠: the board shows $1, 2, \dots, 10$ ⁠. Add them by pairing: what is $S_{10}$ ⁠? Answer, then move Add the terms to the running totals.

Correct answer
55 ± 0
Setup, choices and feedback
The Scene at this step
Difference d
1fixed
Ratio q
0.5defaulthidden
Terms n
10fixed
Add the terms
Terms
Accepted within
± 0
Computed from the Scene at this step’s setting
The arithmetic running total Sₙ = n(a₁ + aₙ)/2/nodes/stairs/outputs/arithmeticSum
Wrong answers it recognises
- 110 ± 0
  Misconception: In Gauss’s pairing, forgets to halve: counts every pair twice.
  Feedback from the catalogue
  When you add the sequence to itself written backwards, every pair gives $a_1 + a_n$ ⁠ and there are $n$ ⁠ pairs, but every term has been counted twice: once forwards and once backwards. So the sum is half of that: $S_n = \frac{n(a_1 + a_n)}{2}$ ⁠.
- 50 ± 0
  Misconception: In Gauss’s pairing, forgets to halve: counts every pair twice.
  Feedback
  $\frac{10 \cdot 10}{2} = 50$ ⁠ pairs the last term with nothing. Each pair is the first term plus the last: $1 + 10 = 11$ ⁠.
Feedback for a right answer
$S_{10} = \frac{10(1 + 10)}{2} = 55$ ⁠. Five pairs, $1 + 10$ ⁠, $2 + 9$ ⁠ and so on, each $11$ ⁠.

Hint
$S_n = \frac{n(a_1 + a_n)}{2}$ ⁠.

Worked solution
1. Write the sequence forwards and, under it, backwards: each column adds up to $1 + 10 = 11$ ⁠.
2. There are $10$ ⁠ columns, so the two rows together make $10 \cdot 11 = 110$ ⁠.
3. Every term was counted twice, so $S_{10} = \frac{110}{2} = 55$ ⁠.
If you got $110$ ⁠, you forgot to divide by $2$ ⁠.
Step 5 of 5
Open this step
Odd numbers and squares
Practice · Reach the target
Question
The sequence is the odd numbers, $a_1 = 1$ ⁠ and $d = 2$ ⁠, and the board shows the running totals. Change Terms n so that the sum is $S_n = 49$ ⁠, then press Check.

Goal
- 49 ± 0.01
  At the start of the step: 9
Setup, choices and feedback
The Scene at this step
Difference d
2fixed
Ratio q
0.5defaulthidden
Terms n
3
Add the terms
Running totalsfixed
Read from the Scene at
The arithmetic running total Sₙ = n(a₁ + aₙ)/2/nodes/stairs/outputs/arithmeticSum
One setting that reaches it
Terms n
7
Feedback for a right answer
$1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 = 7^2$ ⁠. The sum of the first $n$ ⁠ odd numbers is always $n^2$ ⁠: $a_n = 2n - 1$ ⁠, so $S_n = \frac{n(1 + 2n - 1)}{2} = n^2$ ⁠.

Feedback for a wrong answer
Not yet. Look at the totals so far: $1, 4, 9, 16, \dots$ ⁠ What are they?

Worked solution
1. $a_n = 1 + (n - 1) \cdot 2 = 2n - 1$ ⁠.
2. $S_n = \frac{n(a_1 + a_n)}{2} = \frac{n \cdot 2n}{2} = n^2$ ⁠.
3. $n^2 = 49$ ⁠, so $n = 7$ ⁠.
4. Set Terms n to $7$ ⁠.

Lesson 2 of 4 · About 20 minutes

Geometric sequences

A sequence that multiplies by the same number at every step: the general term, how to find the ratio, the sum of the first n terms, and why in the end it overtakes any arithmetic sequence.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Multiply by the ratio
Practice · Predict, then look
Question
$a_1 = 1$ ⁠ and $q = 2$ ⁠. What are the first six terms?

Correct answer
- Correct $1, 2, 4, 8, 16, 32$ ⁠
Setup, choices and feedback
The Scene at this step
Difference d
1defaulthidden
Ratio q
2fixed
Terms n
1
Add the terms
Termsfixed
Choices
- Correct $1, 2, 4, 8, 16, 32$ ⁠
- Wrong $1, 3, 5, 7, 9, 11$ ⁠
  Misconception: Treats a geometric sequence as adding the ratio q each step, instead of multiplying by it.
  Feedback from the catalogue
  In a geometric sequence you multiply by the ratio $q$ ⁠ at every step; you do not add it. With $q = 2$ ⁠ the sequence is $1, 2, 4, 8, \dots$ ⁠: each term is twice the one before. Adding a fixed number makes an arithmetic sequence, which grows in a straight line, not a curve.
- Wrong $2, 4, 8, 16, 32, 64$ ⁠
  Misconception: Counts n steps to the nth term instead of n − 1, so the general term comes out one step too far.
  Feedback from the catalogue
  From the first term to the $n$ ⁠th there are only $n - 1$ ⁠ steps: the first term itself is zero steps along. So $a_n = a_1 + (n - 1)d$ ⁠ for an arithmetic sequence and $a_n = a_1 q^{n-1}$ ⁠ for a geometric one. Check with $n = 1$ ⁠: the formula must give $a_1$ ⁠.
- Wrong $2, 4, 6, 8, 10, 12$ ⁠
  Feedback
  Those are the multiples of $2$ ⁠, which grow by adding. Here each term is $2$ ⁠ times the one before.
Once the prediction is locked in, the Scene changes
Terms n
6
The Scene then shows: 32 · The last geometric term shown, aₙ = a₁qⁿ⁻¹/nodes/stairs/outputs/geometricLast
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$1, 2, 4, 8, 16, 32$ ⁠: each term is twice the one before, $a_n = 2^{n-1}$ ⁠. The gaps between terms grow too, so the line through the column tops curves.

Worked solution
1. $a_1 = 1$ ⁠.
2. Multiply by $q = 2$ ⁠ at every step: $2$ ⁠, $4$ ⁠, $8$ ⁠, $16$ ⁠, $32$ ⁠.
3. As a formula: $a_n = a_1 q^{n-1}$ ⁠, so $a_6 = 2^5 = 32$ ⁠.
If you chose $1, 3, 5, \dots$ ⁠, you added $2$ ⁠ instead of multiplying.
Step 2 of 5
Open this step
The general term and the ratio
Practice · Reach the target
Question
Here the first term is $a_1 = 4$ ⁠, and $n = 3$ ⁠ is locked. Change Ratio q so that the third term is $a_3 = 9$ ⁠, then press Check.

Goal
- 9 ± 0.01
  At the start of the step: 4
Setup, choices and feedback
The Scene at this step
Difference d
1defaulthidden
Ratio q
1
Terms n
3fixed
Add the terms
Termsfixed
Other settings
First/nodes/stairs/props/first = 4
Read from the Scene at
The last geometric term shown, aₙ = a₁qⁿ⁻¹/nodes/stairs/outputs/geometricLast
One setting that reaches it
Ratio q
1.5
Feedback for a right answer
$a_3 = 4q^2 = 9$ ⁠, so $q^2 = \frac94$ ⁠ and $q = 1.5$ ⁠. The sequence is $4, 6, 9$ ⁠: each term is $50\%$ ⁠ more than the one before.

Feedback for a wrong answer
Not yet. From $a_1$ ⁠ to $a_3$ ⁠ you multiply by $q$ ⁠ twice.

Worked solution
1. $a_n = a_1 q^{n-1}$ ⁠, so $a_3 = 4q^2$ ⁠.
2. $4q^2 = 9$ ⁠, so $q^2 = 2.25$ ⁠.
3. $q = \sqrt{2.25} = 1.5$ ⁠.
4. Set Ratio q to $1.5$ ⁠.
If you tried $q = \frac94 = 2.25$ ⁠, you multiplied only once: from $a_1$ ⁠ to $a_3$ ⁠ there are two steps.
Step 3 of 5
Open this step
A sum of powers of 2
Practice · Calculate
Question
$a_1 = 1$ ⁠, $q = 2$ ⁠ and $n = 10$ ⁠. What is $1 + 2 + 4 + \dots + 512$ ⁠? Answer, then move Add the terms to the running totals.

Correct answer
1023 ± 0
Setup, choices and feedback
The Scene at this step
Difference d
1defaulthidden
Ratio q
2fixed
Terms n
10fixed
Add the terms
Terms
Accepted within
± 0
Computed from the Scene at this step’s setting
The geometric running total Sₙ = a₁(qⁿ − 1)/(q − 1)/nodes/stairs/outputs/geometricSum
Wrong answers it recognises
- 1024 ± 0
  Misconception: Thinks 1 + 2 + 4 + … + 2ⁿ⁻¹ equals 2ⁿ, the next term, rather than 2ⁿ − 1.
  Feedback from the catalogue
  $1 + 2 + 4 + \dots + 2^{n-1}$ ⁠ is $1$ ⁠ less than the next term, $2^n$ ⁠: $1 + 2 = 3$ ⁠, $1 + 2 + 4 = 7$ ⁠, $1 + 2 + 4 + 8 = 15$ ⁠. The formula $S_n = \frac{a_1(q^n - 1)}{q - 1}$ ⁠ with $q = 2$ ⁠ gives exactly $2^n - 1$ ⁠.
- 2047 ± 0
  Misconception: Counts n steps to the nth term instead of n − 1, so the general term comes out one step too far.
  Feedback
  $2^{11} - 1 = 2047$ ⁠ is the sum of eleven terms. The tenth term is $2^9 = 512$ ⁠, so the sum is $2^{10} - 1$ ⁠.
Feedback for a right answer
$S_{10} = \frac{2^{10} - 1}{2 - 1} = 1023$ ⁠: one less than the next term, $1024$ ⁠.

Hint
$1 + 2 = 3$ ⁠, $1 + 2 + 4 = 7$ ⁠, $1 + 2 + 4 + 8 = 15$ ⁠. What is the rule?

Worked solution
1. $S_n = \frac{a_1(q^n - 1)}{q - 1}$ ⁠.
2. $S_{10} = \frac{1 \cdot (2^{10} - 1)}{2 - 1} = 2^{10} - 1$ ⁠.
3. $2^{10} = 1024$ ⁠, so $S_{10} = 1023$ ⁠.
If you got $1024$ ⁠, that is the next term, $a_{11}$ ⁠. If you got $512$ ⁠, that is only the last term.
Step 4 of 5
Open this step
A sum of halves
Practice · Calculate
Question
$a_1 = 1$ ⁠, $q = 0.5$ ⁠ and $n = 6$ ⁠: $1 + \frac12 + \frac14 + \dots + \frac{1}{32}$ ⁠. What is $S_6$ ⁠? Answer, then move Add the terms to the running totals.

Correct answer
1.969 ± 0.001
Setup, choices and feedback
The Scene at this step
Difference d
1defaulthidden
Ratio q
0.5fixed
Terms n
6fixed
Add the terms
Terms
Accepted within
± 0.001
Computed from the Scene at this step’s setting
The geometric running total Sₙ = a₁(qⁿ − 1)/(q − 1)/nodes/stairs/outputs/geometricSum
Wrong answers it recognises
- 2 ± 0.0005
  Misconception: Thinks an endless sum is a little less than its limit (0.999… < 1), because no running total reaches it.
  Feedback
  $2$ ⁠ is the endless sum, which the running totals close in on. Six terms do not reach it yet: they are $\frac{1}{32}$ ⁠ short.
Feedback for a right answer
$S_6 = \frac{1 - 0.5^6}{1 - 0.5} = 2\left(1 - \frac{1}{64}\right) = 1.96875$ ⁠. It is close to $2$ ⁠, but short of it by exactly the last term, $\frac{1}{32}$ ⁠.

Worked solution
1. When $q < 1$ ⁠ it is easier to write $S_n = \frac{a_1(1 - q^n)}{1 - q}$ ⁠.
2. $0.5^6 = \frac{1}{64}$ ⁠.
3. $S_6 = \frac{1 - \frac{1}{64}}{0.5} = 2 \cdot \frac{63}{64} = 1.96875$ ⁠.
If you answered $2$ ⁠, that is the sum of all the terms, forever, not of the first six.
Step 5 of 5
Open this step
Which one wins
Practice · Choose
Question
On the left $a_1 = 1$ ⁠ and $d = 3$ ⁠; on the right $a_1 = 1$ ⁠ and $q = 1.5$ ⁠. At $n = 4$ ⁠ the arithmetic sequence is ahead: $a_4 = 10$ ⁠ against $3.375$ ⁠. Which will be bigger at the tenth term?

Correct answer
- Correct The geometric one
Setup, choices and feedback
The Scene at this step
Difference d
3fixed
Ratio q
1.5fixed
Terms n
4
Add the terms
Termsfixed
Choices
Shown in a random order.
- Correct The geometric one
- Wrong The arithmetic one
  Misconception: Expects a sequence that starts ahead by adding to stay ahead of one that multiplies by more than 1.
  Feedback from the catalogue
  An arithmetic sequence adds the same amount at every step, while a geometric one with $q > 1$ ⁠ adds more at each step than at the one before, because what it adds is a fixed share of what is already there. So in the end it overtakes any arithmetic sequence, however far behind it starts.
- Wrong They will be equal
  Feedback
  They do cross between $n = 8$ ⁠ and $n = 9$ ⁠, but only to pass: after that the geometric one keeps pulling away.
Feedback for a right answer
The arithmetic $a_{10}$ ⁠ is $1 + 9 \cdot 3 = 28$ ⁠, and the geometric one is $1.5^9 \approx 38.4$ ⁠. The arithmetic sequence always adds $3$ ⁠; the geometric one adds half of what it has, and that only grows.

Worked solution
1. Arithmetic: $a_{10} = 1 + 9 \cdot 3 = 28$ ⁠.
2. Geometric: $a_{10} = 1.5^9$ ⁠. $1.5^2 = 2.25$ ⁠, $1.5^4 \approx 5.06$ ⁠, $1.5^8 \approx 25.6$ ⁠, and $1.5^9 \approx 38.4$ ⁠.
3. $38.4 > 28$ ⁠: the geometric one is bigger.
4. Move Terms n to see where it overtakes.

Lesson 3 of 4 · About 25 minutes

Infinite series

What happens when you add terms forever: why a half plus a quarter plus an eighth adds up to exactly 2, the formula for the sum of a convergent geometric series, when there is no finite sum, and why 0.999… equals 1.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Halves forever
Practice · Predict, then look
Question
$a_1 = 1$ ⁠ and $q = 0.5$ ⁠. The board shows the running totals of ten terms. If you keep adding forever, $1 + \frac12 + \frac14 + \dots$ ⁠, what is the sum?

Correct answer
- Correct Exactly $2$ ⁠
Setup, choices and feedback
The Scene at this step
Difference d
1defaulthidden
Ratio q
0.5fixed
Terms n
10
Add the terms
Running totalsfixed
Choices
- Correct Exactly $2$ ⁠
- Wrong There is no total: it grows without limit
  Misconception: Thinks adding infinitely many positive terms must give an infinite total.
  Feedback from the catalogue
  Infinitely many positive terms can add up to a finite number if they shrink fast enough. In $1 + \frac12 + \frac14 + \dots$ ⁠ each term fills half of the gap left to $2$ ⁠, so the running totals never pass $2$ ⁠ and come as close to it as you like. The endless sum is exactly $2$ ⁠.
- Wrong A little less than $2$ ⁠
  Misconception: Thinks an endless sum is a little less than its limit (0.999… < 1), because no running total reaches it.
  Feedback from the catalogue
  It is true that no running total reaches the limit. But the endless sum is defined as the number the running totals close in on, so it equals the limit exactly. That is why $0.999\ldots = 1$ ⁠: there is no number between them.
- Wrong $1$ ⁠
  Feedback
  $1$ ⁠ is only the first term. Two terms already make $1.5$ ⁠.
Once the prediction is locked in, the Scene changes
Add the terms
Foreverfixed
The Scene then shows: 2 · The sum of all the geometric terms, a₁/(1 − q), NaN when q ≥ 1/nodes/stairs/outputs/geometricLimit
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Exactly $2$ ⁠. Each term fills half of the gap left to $2$ ⁠, so the running totals come as close to $2$ ⁠ as you like and never pass it. The glass case fills exactly.

Worked solution
1. After $n$ ⁠ terms, $S_n = 2\left(1 - 0.5^n\right)$ ⁠: the gap to $2$ ⁠ is $2 \cdot 0.5^n$ ⁠.
2. As $n$ ⁠ grows, $0.5^n$ ⁠ shrinks to zero, so $S_n$ ⁠ closes in on $2$ ⁠.
3. The endless sum is the number the totals close in on: $S_\infty = \frac{1}{1 - 0.5} = 2$ ⁠.
Step 2 of 5
Open this step
The formula for the endless sum
Practice · Calculate
Question
Here the first term is $a_1 = 3$ ⁠ and $q = 0.75$ ⁠: $3 + 2.25 + 1.6875 + \dots$ ⁠. What is the endless sum? Answer, then move Add the terms to Forever.

Correct answer
12 ± 0.01
Setup, choices and feedback
The Scene at this step
Difference d
1defaulthidden
Ratio q
0.75fixed
Terms n
10
Add the terms
Running totals
Other settings
First/nodes/stairs/props/first = 3
Accepted within
± 0.01
Computed from the Scene at this step’s setting
The sum of all the geometric terms, a₁/(1 − q), NaN when q ≥ 1/nodes/stairs/outputs/geometricLimit
Wrong answers it recognises
- 4 ± 0.01
  Misconception: Uses 1/(1 − q) for an endless geometric sum and forgets to multiply by the first term a₁.
  Feedback from the catalogue
  The formula is $S_\infty = \frac{a_1}{1 - q}$ ⁠. $\frac{1}{1 - q}$ ⁠ is right only when the first term is $1$ ⁠. When every term is three times bigger, so is the sum.
- 9 ± 0.01
  Misconception: Uses 1/(1 − q) for an endless geometric sum and forgets to multiply by the first term a₁.
  Feedback
  $9 = \frac{2.25}{0.25}$ ⁠ is the sum starting from the second term. The sum includes $a_1 = 3$ ⁠ as well.
Feedback for a right answer
$S_\infty = \frac{3}{1 - 0.75} = \frac{3}{0.25} = 12$ ⁠.

Worked solution
1. $|q| < 1$ ⁠, so the sum has a finite value: $S_\infty = \frac{a_1}{1 - q}$ ⁠.
2. $1 - q = 0.25$ ⁠.
3. $S_\infty = \frac{3}{0.25} = 12$ ⁠.
If you got $4$ ⁠, you forgot to multiply by $a_1 = 3$ ⁠.
Step 3 of 5
Open this step
Close to 1
Practice · Reach the target
Question
$a_1 = 1$ ⁠. Change Ratio q so that the endless sum is $10$ ⁠, then press Check.

Goal
- 10 ± 0.01
  At the start of the step: 2
Setup, choices and feedback
The Scene at this step
Difference d
1defaulthidden
Ratio q
0.5
Terms n
10
Add the terms
Forever
Read from the Scene at
The sum of all the geometric terms, a₁/(1 − q), NaN when q ≥ 1/nodes/stairs/outputs/geometricLimit
One setting that reaches it
Ratio q
0.9
Terms n
1
Feedback for a right answer
$\frac{1}{1 - q} = 10$ ⁠, so $1 - q = 0.1$ ⁠ and $q = 0.9$ ⁠. The closer $q$ ⁠ is to $1$ ⁠, the bigger the sum: at $q = 0.99$ ⁠ it is already $100$ ⁠.

Feedback for a wrong answer
Not yet. Which $q$ ⁠ gives $1 - q = \frac{1}{10}$ ⁠?

Worked solution
1. $S_\infty = \frac{a_1}{1 - q} = \frac{1}{1 - q}$ ⁠.
2. $\frac{1}{1 - q} = 10$ ⁠, so $1 - q = 0.1$ ⁠.
3. $q = 0.9$ ⁠. Set Ratio q to $0.9$ ⁠.
Step 4 of 5
Open this step
When there is no total
Practice · Choose
Question
Which of these endless sums have a finite total? Select all that do.

Correct answer
- Correct $1 + 0.5 + 0.25 + \dots$ ⁠, with $q = 0.5$ ⁠
- Correct $1 + 0.9 + 0.81 + \dots$ ⁠, with $q = 0.9$ ⁠
Setup, choices and feedback
The Scene at this step
Difference d
0.5
Ratio q
1
Terms n
10
Add the terms
Forever
Choices
Shown in a random order. More than one choice is correct: every one must be selected.
- Correct $1 + 0.5 + 0.25 + \dots$ ⁠, with $q = 0.5$ ⁠
- Correct $1 + 0.9 + 0.81 + \dots$ ⁠, with $q = 0.9$ ⁠
- Wrong $1 + 1 + 1 + \dots$ ⁠, with $q = 1$ ⁠
  Feedback
  With $q = 1$ ⁠ every term is $1$ ⁠, and $n$ ⁠ terms make $n$ ⁠: the sum grows without limit.
- Wrong $1 + 1.1 + 1.21 + \dots$ ⁠, with $q = 1.1$ ⁠
  Feedback
  With $q > 1$ ⁠ the terms grow, so the sum certainly grows without limit.
- Wrong $1 + 1.5 + 2 + \dots$ ⁠, arithmetic with $d = 0.5$ ⁠
  Feedback
  An endless arithmetic sum has no finite total, except for the sequence whose terms are all $0$ ⁠: the terms never shrink.
Feedback for a right answer
Only $q = 0.5$ ⁠ and $q = 0.9$ ⁠: a geometric series has a finite sum exactly when $-1 < q < 1$ ⁠. In the others the terms do not shrink, and the sum grows without limit.

Worked solution
1. A geometric series has a finite sum when $|q| < 1$ ⁠, and then $S_\infty = \frac{a_1}{1 - q}$ ⁠.
2. $q = 0.5$ ⁠ and $q = 0.9$ ⁠ meet the condition: the sums are $2$ ⁠ and $10$ ⁠.
3. $q = 1$ ⁠ and $q = 1.1$ ⁠ do not: the terms do not shrink.
4. An arithmetic sequence with $d > 0$ ⁠ grows, and so does its sum, without limit.
Step 5 of 5
Open this step
0.999… = 1
Practice · Calculate
Question
$0.999\ldots$ ⁠ is $0.9 + 0.09 + 0.009 + \dots$ ⁠: a geometric series with $a_1 = 0.9$ ⁠ and $q = 0.1$ ⁠. What is its sum? Answer, then move Add the terms to Forever.

Correct answer
1 ± 0.0001
Setup, choices and feedback
The Scene at this step
Difference d
1defaulthidden
Ratio q
0.1fixed
Terms n
4
Add the terms
Running totals
Other settings
First/nodes/stairs/props/first = 0.9
Accepted within
± 0.0001
Computed from the Scene at this step’s setting
The sum of all the geometric terms, a₁/(1 − q), NaN when q ≥ 1/nodes/stairs/outputs/geometricLimit
Wrong answers it recognises
- 0.999 ± 0.00005
  Misconception: Thinks an endless sum is a little less than its limit (0.999… < 1), because no running total reaches it.
  Feedback from the catalogue
  It is true that no running total reaches the limit. But the endless sum is defined as the number the running totals close in on, so it equals the limit exactly. That is why $0.999\ldots = 1$ ⁠: there is no number between them.
Feedback for a right answer
$\frac{0.9}{1 - 0.1} = \frac{0.9}{0.9} = 1$ ⁠. So $0.999\ldots = 1$ ⁠ exactly: two ways of writing the same number.

Worked solution
1. $|q| = 0.1 < 1$ ⁠, so $S_\infty = \frac{a_1}{1 - q}$ ⁠.
2. $S_\infty = \frac{0.9}{0.9} = 1$ ⁠.
If you answered $0.999$ ⁠, that is the sum of the first three terms. The endless sum is the limit, $1$ ⁠.

Lesson 4 of 4 · About 20 minutes

Fibonacci spirals

Why you count 8 and 13 or 34 and 55 spirals on a sunflower or a pine cone: a fixed turn between seeds, fractions of a turn that make spokes, the golden angle that lies near no simple fraction, and the Fibonacci numbers that close in on it.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Spokes
Practice · Predict, then look
Question
Now each seed turns $\frac13$ ⁠ of a turn from the last, and the seeds fall into $3$ ⁠ spokes. How many spokes will they fall into at $\frac38$ ⁠ of a turn, $135^\circ$ ⁠?

Correct answer
- Correct $8$ ⁠
Setup, choices and feedback
The Scene at this step
Turn between seeds
120 °fixed
Plant
Sunflowerfixed
Show the spirals
Offfixed
Choices
- Correct $8$ ⁠
- Wrong $3$ ⁠
  Misconception: Reads the number of spokes from the numerator of the fraction of a turn instead of its denominator.
  Feedback from the catalogue
  When each seed turns $\frac{p}{q}$ ⁠ of a turn, after $q$ ⁠ seeds the total is $p$ ⁠ whole turns and the seed points the same way again. So the seeds fall into $q$ ⁠ spokes: the denominator, not the numerator. At $\frac38$ ⁠ of a turn there are $8$ ⁠ spokes, visited over $3$ ⁠ turns.
- Wrong $11$ ⁠
  Feedback
  $3 + 8 = 11$ ⁠ counts nothing. The seeds point the same way again after $8$ ⁠ seeds, when the total is $3$ ⁠ whole turns.
- Wrong No spokes: the seeds spread out evenly
  Feedback
  $\frac38$ ⁠ is a simple fraction, so every eighth seed points the same way: straight spokes with gaps between them.
Once the prediction is locked in, the Scene changes
Turn between seeds
135 °fixed
The Scene then shows: 8 · Q when the turn is p/q of a full turn (q up to 21, to half the slider's 0.01° step): the seeds then line up in q straight spokes; 0 for a turn near no such fraction, such as the golden angle/nodes/flower/outputs/spokes
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$8$ ⁠ spokes: after $8$ ⁠ seeds the total is $8 \cdot \frac38 = 3$ ⁠ whole turns, and the ninth seed lands behind the first.

Worked solution
1. Seed number $k$ ⁠ has turned $k \cdot \frac38$ ⁠ of a turn.
2. That is a whole number for the first time at $k = 8$ ⁠: $8 \cdot \frac38 = 3$ ⁠.
3. So every eighth seed points the same way, and there are $8$ ⁠ spokes.
If you chose $3$ ⁠, you counted the whole turns, not the spokes.
Step 2 of 5
Open this step
Fibonacci’s fractions
Practice · Reach the target
Question
$\frac13$ ⁠, $\frac25$ ⁠, $\frac38$ ⁠: the numerators and denominators are Fibonacci numbers. The next fraction in the row is $\frac{5}{13}$ ⁠. Change Turn between seeds so that the seeds fall into $13$ ⁠ spokes, then press Check.

Goal
- 13 ± 0.1
  At the start of the step: 8
Setup, choices and feedback
The Scene at this step
Turn between seeds
135 °
Plant
Sunflowerfixed
Show the spirals
Offfixed
Read from the Scene at
Q when the turn is p/q of a full turn (q up to 21, to half the slider's 0.01° step): the seeds then line up in q straight spokes; 0 for a turn near no such fraction, such as the golden angle/nodes/flower/outputs/spokes
One setting that reaches it
Turn between seeds
138.5 °
Feedback for a right answer
$\frac{5}{13}$ ⁠ of a turn is $\frac{5}{13} \cdot 360^\circ \approx 138.46^\circ$ ⁠. The fractions $\frac25$ ⁠, $\frac38$ ⁠, $\frac{5}{13}$ ⁠, $\frac{8}{21}$ ⁠ close in on the golden angle, from above and below in turn.

Feedback for a wrong answer
Not yet. How many degrees are $\frac{5}{13}$ ⁠ of $360^\circ$ ⁠? The angle must be right to a hundredth of a degree.

Hint
$\frac{5}{13} \cdot 360^\circ$ ⁠, to two decimal places.

Worked solution
1. To get $13$ ⁠ spokes, the turn must be $\frac{p}{13}$ ⁠ of a turn.
2. In the slider’s range, $120^\circ$ ⁠ to $150^\circ$ ⁠, the only such fraction is $\frac{5}{13}$ ⁠.
3. $\frac{5}{13} \cdot 360^\circ = \frac{1800^\circ}{13} \approx 138.46^\circ$ ⁠.
4. Set Turn between seeds to $138.46^\circ$ ⁠.
Step 3 of 5
Open this step
The golden angle
Practice · Calculate
Question
The turn is the golden angle, $137.508^\circ$ ⁠, and the golden ratio is $\varphi = \frac{1 + \sqrt5}{2} \approx 1.618$ ⁠. What fraction of a full turn is the golden angle? (Give a decimal.)

Correct answer
0.382 ± 0.001
Setup, choices and feedback
The Scene at this step
Turn between seeds
137.5 °fixed
Plant
Sunflowerfixed
Show the spirals
Offfixed
Accepted within
± 0.001
Computed from the Scene at this step’s setting
The turn between seeds as a fraction of a full turn, angle / 360/nodes/flower/outputs/turnFraction
Wrong answers it recognises
- 0.618 ± 0.001
  Misconception: Takes the golden angle as 1/φ ≈ 0.618 of a turn (222.5°), the larger part, rather than the smaller part 1 − 1/φ.
  Feedback
  $0.618 = \frac{1}{\varphi}$ ⁠ is the larger part of the turn, $222.5^\circ$ ⁠. The golden angle is what is left: $1 - 0.618$ ⁠.
Feedback for a right answer
$\frac{137.508}{360} \approx 0.382 = 1 - \frac{1}{\varphi} = \frac{1}{\varphi^2}$ ⁠. The golden angle splits the turn in the golden ratio: the larger part, $222.5^\circ$ ⁠, divided by the smaller is $\varphi$ ⁠ again.

Worked solution
1. A fraction of a turn is the angle divided by $360^\circ$ ⁠: $\frac{137.508}{360} \approx 0.382$ ⁠.
2. Check it against the golden ratio: $\frac{1}{\varphi} \approx 0.618$ ⁠, and $1 - 0.618 = 0.382$ ⁠.
3. The golden angle is the smaller part of the turn split in the ratio $\varphi$ ⁠.
If you got $0.618$ ⁠, that is the larger part, $222.5^\circ$ ⁠: a turn that way is $137.5^\circ$ ⁠ the other way.
Step 4 of 5
Open this step
Counting spirals
Practice · Choose
Question
Choose On under Show the spirals. Each colour is one family of spirals, and the number beside it counts its arms. What do the numbers have in common?

Correct answer
- Correct They are consecutive Fibonacci numbers: each is the sum of the two before
Setup, choices and feedback
The Scene at this step
Turn between seeds
137.5 °fixed
Plant
Sunflowerfixed
Show the spirals
Off
Choices
Shown in a random order.
- Correct They are consecutive Fibonacci numbers: each is the sum of the two before
- Wrong They are all multiples of the same number
  Feedback
  $34$ ⁠ and $55$ ⁠ (or $21$ ⁠ and $34$ ⁠) share no factor but $1$ ⁠.
- Wrong They are all prime
  Feedback
  $34 = 2 \cdot 17$ ⁠ and $55 = 5 \cdot 11$ ⁠: not prime.
- Wrong Nothing: they depend on the flower
  Feedback
  At the golden angle the counts are always consecutive Fibonacci numbers. Only which ones you see depends on the flower’s size: on a small head, $21$ ⁠ and $34$ ⁠.
Feedback for a right answer
$34$ ⁠, $55$ ⁠ and $89$ ⁠ (on a phone $21$ ⁠, $34$ ⁠ and $55$ ⁠): consecutive Fibonacci numbers, $34 + 55 = 89$ ⁠. They are the denominators of the fractions that close in on the golden angle, so seeds $34$ ⁠ or $55$ ⁠ steps apart sit in nearly the same direction and read as a spiral.

Worked solution
1. Read the numbers beside the spirals: $34$ ⁠, $55$ ⁠ and $89$ ⁠ (on a small screen $21$ ⁠, $34$ ⁠ and $55$ ⁠).
2. $21 + 34 = 55$ ⁠ and $34 + 55 = 89$ ⁠: each is the sum of the two before.
3. They are the denominators of the fractions $\frac{13}{34}$ ⁠, $\frac{21}{55}$ ⁠, $\frac{34}{89}$ ⁠, which close in on $0.382$ ⁠, the golden angle.
Step 5 of 5
Open this step
A pine cone
Practice · Predict, then look
Question
Switch Plant to the pine cone, at the same golden angle. A pine cone has only $180$ ⁠ scales. How many spirals will the two families have?

Correct answer
- Correct $8$ ⁠ and $13$ ⁠
Setup, choices and feedback
The Scene at this step
Turn between seeds
137.5 °fixed
Plant
Sunflower
Show the spirals
On
Choices
- Correct $8$ ⁠ and $13$ ⁠
- Wrong $34$ ⁠ and $55$ ⁠, as on the sunflower
  Feedback
  With few scales you see the short spirals: each scale’s nearest neighbours are $8$ ⁠ and $13$ ⁠ steps from it.
- Wrong $10$ ⁠ and $20$ ⁠
  Feedback
  The angle has not changed, so the counts stay Fibonacci numbers.
- Wrong A pine cone has no spirals
  Feedback
  The same rule, the same turn for every scale, makes spirals on a pine cone too.
Once the prediction is locked in, the Scene changes
Plant
Pine cone
Show the spirals
On
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$8$ ⁠ and $13$ ⁠, consecutive Fibonacci numbers again. Real pine cones count the same way: most have $8$ ⁠ and $13$ ⁠, some $5$ ⁠ and $8$ ⁠.

Worked solution
1. The rule has not changed: each scale turns by the golden angle.
2. A scale’s nearest neighbours are the ones a Fibonacci number of steps away.
3. With a couple of hundred scales those are $8$ ⁠ and $13$ ⁠: smaller numbers than among the sunflower’s thousand seeds.

Across the unit · 10 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Arithmetic sequences

Add the difference

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The general term

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Terms far away

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Adding in pairs

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Odd numbers and squares

Question

Goal

The Scene at this step

One setting that reaches it

Worked solution

Geometric sequences

Multiply by the ratio

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The general term and the ratio

Question

Goal

The Scene at this step

One setting that reaches it

Worked solution

A sum of powers of 2

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A sum of halves

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

Which one wins

Question

Correct answer

The Scene at this step

Choices

Worked solution

Infinite series

Halves forever

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The formula for the endless sum

Question

Correct answer

The Scene at this step