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]
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.
