X Induction (Series & Divisibility)

Question 1

process for series

Answer 1

Step 1: To prove the statement is true for the base case ( ie
n = 0, n = 1)

sub n = base case
LHS =
RHS =
LHS = RHS
∴ the statement is true for the base case ( ie n = 0, n = 1)

Step 2: Assume the statement is true for n = k, where k is a positive integer
∴ [insert series + k] = [insert sum with k subbed in] ☆

Step 3: To prove that the statement is true for n = k + 1
To prove [insert series + k + (k + 1)] = [insert sum with (k + 1) subbed in]

LHS = (do algebra) and use ☆ to arrive to [insert sum with (k + 1) subbed in]
∴ LHS = RHS
∴ the statement is true for n = k + 1
Hence by principle of mathematical induction, the statement is true for all positive integers n

Question 2

process for divisibility

Answer 2

Q is the divider number

Step 1: To prove the statement is true for base case (ie n = 0)

sub n = base case
LHS = XYZ which is divisible by Q
∴ the statement is true for the base case (ie n = 0)

Step 2: Assume the statement is true for n = k, where k is a positive integer > 0 (base case)

sub n = k
XYZ is divisible by Q
∴ XYZ = Qm, for some integer m ☆

Step 3: To prove the statement is true for n = k +1
To prove [XYZ with (k + 1) subbed in] is divisible by Q

(do some algebra with n = k + 1 subbed in) using ☆ to arrive to any integer in brackets ( ie 11m - 2 x 5ᵏ) multiplied by Q

= Q(11m - 2 x 5ᵏ) which is divisible by Q

∴ the statement is true for n = k + 1
Hence by mathematical induction, the statement is true for all positive integers n

Question 3

atomi techniques

Answer 3

Question 4

atomi techniques

Answer 4

Question 5

atomi techniques

Answer 5

Question 6

little exam trick
instead of asking for n = all positive integers
they may ask for n = all odd positive integers

Answer 6

for all odd integers
inductive step to assume for
n = is k+2 instead of k+1

Question 7

general info

Answer 7

Question 8

general sigma notation

Answer 8