Demonstration 1 of 4
Count the additions
Stack A starts at 100 and adds 10. How do you jump straight to a_4?
Each bar is the start of 100 plus n tens. The factor n in the rule counts the additions since index 0, so the teal part grows by one 10 per step.
\[a_n=a_0+nd\]
\[a_4 = 100 + 10(4) = 140\]
a_n is the term with index n; the subscript names a position, it is not multiplication. a_0 is the starting term and d is the constant difference, here 10.
Predict first. Choose index 0. How many 10s have been added, and what does the direct rule give?
Choose an example
Constructed example: the chapter's Stack A, 100, 110, 120, 130.
Calculated values
- Index
- 4
- Additions since a_0
- 4
- Direct rule
- a_4 = 100 + 10(4) = 140
- Recursive check
- 100, 110, 120, 130, 140 (add 10 each step)
- Difference d
- 10 at every step
a_4 = 100 + 10(4) = 100 + 40 = 140. Reaching index 4 takes 4 steps of 10, so 4(10) = 40 has been added to the start of 100. The recursive rule a_(n+1) = a_n + 10 walks there one step at a time and agrees.
Use the idea
For any steady add-the-same-amount pattern, write the start and the step, then multiply the step by the number of steps instead of listing every term.
Where the conclusion applies
The index starts at 0 and every step adds exactly 10. If one difference changed, the list would not be arithmetic over that stretch, and the rule would fail there.
Check your understanding: Using a_n = 100 + 10n, what is a_9?
Chapter 25 source: section "Constant difference means arithmetic". Demonstration C25-D01.
Demonstration 2 of 4
Add 10 or multiply by 1.10
Stacks A and B both begin 100, 110. Why do they part ways?
Stack A adds the same 10 every step, so it climbs in a straight line. Stack B adds a tenth of a growing amount, so its steps grow: 10, 11, 12.1, 13.31, and on.
\[b_n = 100(1.10)^n\]
\[b_4 = 100(1.10)^4 = 146.41\]
b_n is Stack B's term with index n. The multiplier 1.10 scales the whole current term; the exponent n counts how many times it has been applied since b_0 = 100.
Predict first. Move the index to 10. Will Stack B be ahead of Stack A by more than 50?
Choose an example
Constructed example: the chapter's Stacks A and B, extended to later indices.
Calculated values
- Stack A, a_n = 100 + 10n
- a_4 = 100 + 10(4) = 140
- Stack B, b_n = 100(1.10)^n
- b_4 = 100(1.10)^4 = 146.41
- B minus A
- 6.41
- Step added by B at the end
- 13.31
- Step added by A at every step
- 10
a_4 = 100 + 10(4) = 140, and b_4 = 100(1.10)^4 = 146.41. Stack A adds 10 every time. Stack B multiplies the whole current amount by 1.10, so its last step added 13.31. The first few cards look close; the rules are not.
Use the idea
When two rules agree on the first few values, compute one value far out with each rule before trusting either one.
Where the conclusion applies
Both rules are declared, not discovered: four cards alone do not prove how a pattern continues. The far-out values follow only if each rule really keeps going. Values that do not end within six decimal places are rounded and marked about.
Check your understanding: What is b_5 for Stack B?
Chapter 25 source: section "Constant multiplier means geometric". Demonstration C25-D02.
Demonstration 3 of 4
Where does counting start?
The list 7, 11, 15, 19 never changes. Why can its formula change?
Switching the convention relabels every card by one. The formula shifts to match, so each card's value stays put while its index moves.
\[a_n=7+4n\]
\[a_n=7+4(n-1)\]
The index is the label on each card. Starting at a_0 or at a_1 is a choice that must be stated; n - 1 counts the additions when counting starts at 1.
Predict first. Pick the 4th entry under each convention. Do the two rules give the same value?
Choose an example
Constructed example: the chapter's list 7, 11, 15, 19 under both index conventions.
Calculated values
- Convention
- first entry is a_0
- Direct rule
- a_n = 7 + 4n
- Index of this entry
- 3
- Value
- a_3 = 7 + 4(3) = 19
- Additions of 4 since the first entry
- 3
With the first entry called a_0, entry 4 is a_3: a_3 = 7 + 4(3) = 19. That is 3 additions of 4 after the starting 7. The other convention gives the same card a different index, 4, and the matching formula still lands on 19.
Use the idea
Write the first index beside the first value before using any direct rule. It is the cheapest way to avoid an off-by-one error.
Where the conclusion applies
A four-term arithmetic list with difference 4. The rule with n - 1 holds for n of at least 1; using it with the a_0 labels would put every answer one step behind.
Check your understanding: With the first entry called a_1, what is a_6?
Chapter 25 source: section "Worked example: index zero and index one describe different starting choices". Demonstration C25-D03.
Demonstration 4 of 4
Ten percent keeps the whole
A balance of 100 grows by 10% each round. What do you multiply by?
Each bar is the previous bar times the multiplier. A multiplier above 1 grows the amount, one between 0 and 1 shrinks it, and 0.10 keeps only a tenth.
\[\text{new} = \text{previous} + 0.10(\text{previous})\]
\[100(1.10)^3 = 133.1\]
previous is the amount before a round and new is the amount after it. 0.10(previous) is ten percent of it. Keeping the whole and adding the tenth is multiplying by 1.10.
Predict first. Choose the mistake, multiply by 0.10. What is left after one round?
Choose an example
Constructed example: the chapter's 100-unit token rounds, plus its 20% examples.
Calculated values
- Multiplier per round
- 1.10
- Rounds
- 100 x 1.10 = 110; 110 x 1.10 = 121; 121 x 1.10 = 133.1
- Direct rule
- 100(1.10)^3 = 133.1
- After three rounds
- 133.1 units
100 x 1.10 = 110; 110 x 1.10 = 121; 121 x 1.10 = 133.1. Direct check: 100(1.10)^3 = 133.1. Each round keeps the whole and adds a tenth: 100 + 10 = 110, 110 + 11 = 121, 121 + 12.1 = 133.1.
Use the idea
Turn any percent change into a multiplier first: an increase of p percent keeps the whole and adds p hundredths; a decrease keeps what is left.
Where the conclusion applies
The same percent applies each round to the amount at the start of that round, starting from 100 units. If the percent were taken from the original 100 each time, the pattern would be arithmetic instead.
Check your understanding: A 20% decrease is applied twice to 100. What remains?
Chapter 25 source: section "Ten percent growth keeps the whole". Demonstration C25-D04.