# Explaining the Principle of mathematical induction

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Introduction

1Explaining the Principle of mathematical induction:

Formally the principle of a proof by induction can be stated as follows:

A proposition P (n) involving a positive integer n, is true for all positive integral values of n if, P (1), and P (k) → P (k +1) is true.

This can be explained using a staircase as a simple analogy. Image the proposition that a man can climb a given uniform staircase, to prove this statement we need to show two things. These are that the man can get onto the first step and that he is able to climb from one step to an other. Now relating this to the formal principle of induction, the staircase can be considered the general proposition P (n). The first step of the staircase is P (1), the second P (2), the third P (3), and so on. If we can show that the man can get onto the first step P (1) then we have ironically finished the first step of proving the proposition.

Middle

2 +

3 +

4 +

5 +

6 +

7 +

8 +

9 +

10

+

+

+

+

+

+

+

+

+

+

10

+ 9

+ 8

+ 7

+ 6

+ 5

+ 4

+ 3

+ 2

+ 1

=

=

=

=

=

=

=

=

=

=

Total

11

11

11

11

11

11

11

11

11

11

Sum of sequence = (10 ×11)/ 2 = 55

Term

u1

+……………………………………………………………+

un

+

un

+……………………………………………………………+

u1

=

=

un + u1

Sum of sequence = (n ( un + u1)) / 2

u1 + un

- Find the value (in terms of n) of: 1 + 2 + 4 + 5 + 7 + 8 +… (3n - 2) + (3n - 1)

A value for the above sequence can be derived from the formula for the sum of an arithmetic sequence, which is as follows:

u1 = 1 + 2 = 3

un = (3n - 2) + (3n - 1) = (6n – 3)

(u1 + un) = 3 + (6n – 3) = 6n

Resulting value in terms of n is:

or 6n2

2

8 Consider the table: 1 = 1

2 + 3 + 4 = 9

5 + 6 + 7 + 8 + 9 = 35

Write down the numbers on the right-hand side as the sum of two cubes and then write down the next line of the table.

Conclusion

S1 13 = 1 = 12

S2 13 + 23 = 9 = 32

S3 13 + 23 + 33 = 36 = 62 S4 13 + 23 + 33 + 43 = 100 = 102

S5 13 + 23 + 33 + 43 + 53 = 225 = 152

After writing down the sums of several partial sums I found that the resulting sums where square numbers. Upon taking the square root of the square numbers, I came across another discovery, which was that the resulting numbers were found to be the sum of the sequences on the LHS without the exponents. From this knowledge I can make the conjecture that the sum of n terms is the square of the sum of the sequence disregarding the exponents. The formula for the conjecture lies below:

Where n∈ N

Proving the conjecture

Step 1: Prove true for n = 1

1 = 1

→ LHS=RHS.·. Formula holds

Step 2: Assume true for n = k

Now we have to prove true for n = k + 1

LHS

→ LHS =RHS.·. Formula holds

10 Given the Matrix A = , find the value of A2, A3, A4, A5... and make conjecture for An. Also, prove the conjecture by induction.

A1= , A2 = , A3 = , A3 = , A3 =

Conjecture:

Proving the conjecture:

Step 1: Prove true for n = 1

.·. Formula holds

Step 2: Assume true for n = k

Now we have to prove true for n = k + 1

LHS:

Ak+1 = Ak × A

→ LHS = RHS.·. Formula holds

This student written piece of work is one of many that can be found in our GCSE Consecutive Numbers section.

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