Prove a recurrence pattern
What makes a step from k to k+1 work for every k?
Entering midway? A recurrence defines a new term from earlier terms. Induction links a base case to a general next-case argument.
Start with a concrete question
Accumulate odd or even increments from S(0)=0 and test the proposed formula S(n)=n².
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What do you expect?
Predict before calculating: What makes a step from k to k+1 work for every k?
Change one thing. Inspect what happens.
- Trace the first four partial sums.
- Switch to even increments and locate the first failed square claim.
- Write the base case and the symbolic k→k+1 implication.
- Explain why showing one numerical transition is not an induction proof.
Your settingsMechanism
Calculation belowEvidence
Compare outcomes
Accumulated recurrence versus the proposed square
Horizontal: integer step k. Vertical: sum / proposed value (units).
- 1. Actual accumulated recurrence (solid line)
- 2. Proposed k² (dashed line)
Axes use the displayed ranges.
Read plotted values as a table
| Series | integer step k | sum / proposed value (units) |
|---|---|---|
| Actual accumulated recurrence | 0 | 0 |
| Actual accumulated recurrence | 1 | 1 |
| Actual accumulated recurrence | 2 | 4 |
| Actual accumulated recurrence | 3 | 9 |
| Actual accumulated recurrence | 4 | 16 |
| Proposed k² | 0 | 0 |
| Proposed k² | 1 | 1 |
| Proposed k² | 2 | 4 |
| Proposed k² | 3 | 9 |
| Proposed k² | 4 | 16 |
Follow the calculation
- Base case S(0)=0=0².
- S(1)=S(0)+1=1; proposed 1²=1.
- S(2)=S(1)+3=4; proposed 2²=4.
- S(3)=S(2)+5=9; proposed 3²=9.
- S(4)=S(3)+7=16; proposed 4²=16.
- Induction step: if S(k)=k², then S(k+1)=S(k)+2(k+1)−1=k²+2k+1=(k+1)².
- The base case plus a valid implication from an arbitrary k to k+1 establishes the claim for every nonnegative integer. Checking one selected step is not that general implication.
- The plotted points compare accumulated sums with the proposed square on integer steps. Joining lines guide the eye; no fractional recurrence steps are computed. Agreement on this finite plot is not an induction proof.
A proof by induction has two obligations: a starting case and a general step. Edit the supplied argument, including its domain and why the next case follows.
Repair the induction proof
Write the base case, explicitly state the induction hypothesis for arbitrary k≥0, derive the next case and conclude the domain. Explain where the even-increment recurrence breaks the argument.
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Compare with my earlier draft
It works at n=1. Assume the formula is true. Therefore it is true at every n.
Explain what you changed and why in your observation notes. Text differences do not establish quality.
Inspect an example solution after your attempt
S(0)=0=0². Suppose S(k)=k² for an arbitrary nonnegative k. Then S(k+1)=S(k)+2(k+1)−1=k²+2k+1=(k+1)². The base and step establish all n≥0. Even increments add 2k+2 instead, leaving an extra 1 in this step.
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Explain the mechanism
Prove a recurrence pattern
Accumulate odd or even increments from S(0)=0 and test the proposed formula S(n)=n².
Concepts to recall
A recurrence defines a new term from earlier terms. Induction links a base case to a general next-case argument.
Understand the mechanism
The first odd integers sum as 1,4,9,16.
Assume the statement for an arbitrary k: S(k)=k². The next increment is 2(k+1)−1=2k+1, giving S(k+1)=(k+1)².
The assumption is conditional for an arbitrary step, not permission to assume the conclusion for every n. Base plus step carries the claim along the nonnegative integers.
Symbols and calculation
S(0)=0; S(n)=S(n−1)+(2n−1). Induction: S(k)=k² ⇒ S(k+1)=k²+2k+1=(k+1)².
Worked example
Base case S(0)=0=0².
S(1)=S(0)+1=1; proposed 1²=1.
S(2)=S(1)+3=4; proposed 2²=4.
S(3)=S(2)+5=9; proposed 3²=9.
S(4)=S(3)+7=16; proposed 4²=16.
Induction step: if S(k)=k², then S(k+1)=S(k)+2(k+1)−1=k²+2k+1=(k+1)².
The base case plus a valid implication from an arbitrary k to k+1 establishes the claim for every nonnegative integer. Checking one selected step is not that general implication.
The plotted points compare accumulated sums with the proposed square on integer steps. Joining lines guide the eye; no fractional recurrence steps are computed. Agreement on this finite plot is not an induction proof.
Your investigation
Trace the first four partial sums.
Switch to even increments and locate the first failed square claim.
Write the base case and the symbolic k→k+1 implication.
Explain why showing one numerical transition is not an induction proof.
Limits and interpretation
- These are bounded teaching models with explicit assumptions, not evidence of a real-world intervention.
Try a fresh case
What is the sum of the first seven positive odd integers?
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Take the investigation further
Choose any extension. These are project prompts for you to carry out; they are not automatically assessed.
School investigation
Explain a fresh small numerical case with a diagram or table and an argument.
College investigation
Implement the method, test a boundary or counterexample, and justify its assumptions.
Workplace investigation
Use a small decision scenario, state units and constraints, and test how the recommendation changes.
Keep the evidence, choose your next step
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