## PROOF BY MATHEMATICAL INDUCTION

Step1: show that the statement is true for the initial value.

Step 2: assume that the statement is true for n = k

Step 3: prove that the statement is true for n = k+1

Then it follows the statement must be true for all values of n

Question 1:

Prove 2+4+6+………….2n = n[n+1] by mathematical induction.

When n = 1

LHS 2n RHS= n[n+1]

2[1] = 1[1+1]

2 = 1[2]

N = k

2= 2

Show n is true for n = k+1

That is 2+4+6+………+2k+2[k+1}

= [k+1]{k+1+1]

2+4+6……..2k+2[k+1][k+2]

Recall that 2+4+6…..2k = k[k+1]+2[k+1]

Therefore, it is true for n = k+1

So therefore, the assumption is true n≥1

**MATHEMATICALINDUCTION 2**

Base case: show n=1 is true

Inductive hypothesis: assume that n=k is true

Inductive step: shoe that n= k+1 is true

Prove that -1 is divisible by 7 for all positive integers n

Sn : -1 = 7m for some integer of m

BASE CASE [ show that n=1 is true]

-1 = 7[1]

−1 = 7

]

**INDUCTIVE HYPOTHESIS** [assume n=k is true]

Assume that Sn : is true for some integer of m ………….eqn 1

INDUCTIVE STEP [show that n = k+1 is true]

……………………….eqn 2

Since from equation 1

```
Substitute that for in equation 2
= [7m+1]8-1
=56m+8-1
= 56m+7
= 7[8m+1]
```

8m+1is an integer by closure, so let 8m+1 = p for some integer of p

This implies that 7 divides ,since the statement holds for n = 1 and true for n=k implies that n=k+1 is also true, **the statement is true for all positive integer.**