1. 2.4 Mathematical Induction

2. Prove that a given statement is true
LHS RHS Statement …..+ n = n(n + 1)/2 1. Basis Step: Replace n with 1 1 = 1(1 +1)/2 is this true? 1 = 1(2)/2 yes it is true…continue to the next step.

3. 2. Induction step a. Replace n with k LHS RHS 1 + 2 + 3……….+k =
b. Replace k with k+1
LHS RHS
…………… = (k+1)((k+1) + 1)/2
Solve LHS RHS
(k + 1)(k + 2)/2
k(k + 1)/2
k + 1
k + 1
k(k + 1)/2

4. LHS
(k+ 1)/1 * 2/2 = 2(k+1)/2
2(k + 1 )/2 = 2k + 2/2
k(k+1)/2 = k2 + k/2
2k +2/2 + k2 + k/2 = 2k k2 + k/2
= k2+ 3k + 2/2
= (k + 1)(k + 2)/2
The left hand side is equal to the right hand side.

5. FOIL (First Outer Inner Last)
(k + 1) (k + 2)
First (k) (k) = k2
Outer = (k) (2) = 2k
Inner = (1) (k) = k
Last = (1) (2) = 2
k2+ 2k + k + 2 = k2 + 3k + 2

6. Factor the expression
k2 + 3k + (k + 1) (k + 2) What 2 numbers when added = 3? = 3 Can you multiply the 2 numbers to = 2? Yes 2

7. Normally, when you find that the basis step is false, you don’t need to go any further with the problem. It is possible that the induction step could be found true when the basis step is found to be false.