Mathematical Induction

Prove a statement for every whole number at once.

12 problems · about 22 minutes · TEKS P.5.E

What this lesson teaches

The student proves summation formulas using mathematical induction.

Induction proves a statement for every whole number using two parts.
The base case starts the chain; the inductive step passes it along.
Assuming n = k to prove n = k + 1 is a conditional argument, not circular reasoning.

Straightforward practice. Get the method working first.

5 problems

n(n+1)/2 at n = 8?
Answer 36
Why 36.
Why is a base case necessary?
Answer Without a true first case, the chain of implications never starts.
Why Nothing topples if the first domino never falls.
n(n+1)/2 at n = 6?
Answer 21
Why 6 × 7 ÷ 2.
n(n+1)/2 at n = 10?
Answer 55
Why 10 × 11 ÷ 2.
The sum of the first n odd numbers is n². What is it for n = 5?
Answer 25
Why 1 + 3 + 5 + 7 + 9.

The same ideas with more to keep track of.

3 problems

How many parts does a proof by induction have?
Answer 2
Why The base and the step.
In the inductive step, which case do you assume? Enter the value as a number of steps past k, so 0 means k itself.
Answer 0
Why You assume n = k and prove n = k + 1.
The sum of the first n odd numbers at n = 8?
Answer 64
Why 8².

Mixed problems. Work out what kind of question it is before you start.

4 problems

Put a proof by induction in order.
Answer 1. State the formula to be proved. 2. Verify the base case, usually n = 1. 3. Assume the formula holds for n = k. 4. Derive the formula for n = k + 1.
Why The base case is checked before any assumption is made.
n(n+1)/2 at n = 100?
Answer 5050
Why 100 × 101 ÷ 2.
The Base: For 1 + 2 + ... + n = n(n+1)/2, what does the right side give when n = 4?
Answer 10
Why 10, matching 1 + 2 + 3 + 4.
The Missing Part: A proof verifies n = 1 but never proves the step from k to k + 1. Is it complete? Enter 1 for yes, 0 for no.
Answer 0
Why No. Both parts are required.