3. Proof by Induction

Video coming soon2 worked examples

Theory

Proof by induction establishes a statement $P(n)$ for all positive integers $n$ in three parts:

Base case: show $P(1)$ is true.
Inductive step: assume $P(k)$ is true (the inductive hypothesis), and use it to prove $P(k+1)$ .
Conclusion: state that, by induction, $P(n)$ holds for all $n \geq 1$ .

The Golden Rule: you must actually use the inductive hypothesis when proving $P(k+1)$ — that is the heart of the method — and never skip the base case or the concluding statement.

⚠️ Common Examiner Traps

Missing the base case: without $P(1)$ the induction has no foundation.
Not using the hypothesis: the $P(k+1)$ working must build on the assumed $P(k)$ .
No conclusion: finish with the “therefore true for all $n$ ” statement to earn the final mark.

Worked examples

Example 1

Prove by induction that $\displaystyle\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ for all positive integers $n$ .

Base case: for $n=1$ , the left side is $1$ and the right side is $\frac{1(2)}{2} = 1$ . True.

Inductive step: assume the result holds for $n = k$ , i.e. $\sum_{r=1}^{k} r = \frac{k(k+1)}{2}$ . Then:

\sum_{r=1}^{k+1} r = \frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}

Conclusion: this is the formula with $n = k+1$ , so if it holds for $k$ it holds for $k+1$ . Since it holds for $n=1$ , by induction it holds for all $n \geq 1$ .

Example 2

Prove by induction that $8^n - 1$ is divisible by $7$ for all positive integers $n$ .

Base case: for $n=1$ , $8^1 - 1 = 7$ , which is divisible by $7$ . True.

Inductive step: assume $8^k - 1$ is divisible by $7$ , so $8^k = 7m + 1$ for some integer $m$ . Then:

8^{k+1} - 1 = 8 \cdot 8^k - 1 = 8(7m + 1) - 1 = 56m + 7 = 7(8m + 1)

Conclusion: this is a multiple of $7$ , so the result holds for $k+1$ . Since it holds for $n=1$ , by induction $8^n - 1$ is divisible by $7$ for all $n \geq 1$ .