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  • Level: GCSE
  • Subject: Maths
  • Document length: 1276 words

Explaining the Principle of mathematical induction

Extracts from this document...

Introduction

1 Explaining 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. The second step would be to prove that he can get from one step to an other formally put P (k) ? P (k +1). If we can show this than it follows that man can get from the first to the second step, second to the third,... n steps. Thus it can be said that P (n) is true for all positive integral values of n. ...read more.

Middle

A example of this can be can be seen in the table below where the first 10 terms of the sequence given in question 5 have been taken. When adding each pair it can be seen that a consistent sum of 11 is given for each. Since there are 10 pairs we need to multiply by 10 to get the sum , however this is the sum for both of the sequences thus to get the sum for one we need to divide by 2. This principle can be translated into a general formula for all arithmetic sequences, this can be seen in the bottom half of the table . If we know the first term u1 and the last term un then we can find the sum of the pair which will be equal for all other pairs in the sequence. All we need to do now is to multiply this by the number of pairs (number of terms n), and then divide by 2 to the sum for one sequence. Term u1 u2 u3 u3 u5 u6 u7 u8 u9 u10 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + + + + + + + + + + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = = = = = = = = = = ...read more.

Conclusion

= 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 ...read more.

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