Example 1

$$ \sum^n_{i=1}i=\frac{{n(n+1)}}{2} $$

1. Base case: n=1, solve -> 1=1, true!

2. Assume that the formula holds for n=k, so we now have: $$ \sum^k_{i=1}i={\frac{k(k+1)}{2}} $$

3. Induction step, if n=k holds true, then n=k+1 must hold true too, so lets now check for that: $$ \frac{{(k+1)((k+{1}+1))}}{2}= $$

Strong vs simple proof by induction?

• Esentially the same, however he will test on their differeences.

Example 2

S = 1 + a + a² + ... + aⁿ, 0 < a < 1 How would we solve this? 1) We can take aS = a + a² + … + a^n + 1 2) So we have $(1-a)S=1,S=\frac{1}{1-a}$ 3) Hence it must converge to some 1/0…