Help with induction... (1 Viewer)

[wccchick]

Just quicky - can anyone give me a hand with this induction?

use the method of mathematical induction to prove that 4 (to the power of n) + 14 is a multiple of 6 for all n> or equal to 1

 

[insert-username]

To prove: 4ⁿ + 14 = 6P, where P is an integer, for all n≥1 (note that all '.' mean multiplication)

Step 1: Prove true for smallest n

4¹ + 14 = 18

= 6P (where P = 3)

Therefore true for n = 1


Step 2 - Assume true for n = k

4^k + 14 = 6P, where P is an integer


Step 3 - Prove true for n = k+1

4^k+1 + 14 = 6Q, where Q is an integer

LHS = 4.4^k + 14 (if we add one to the index, we're multiplying it again by that number)

= 3.4^k + 4^k + 14

But (4^k + 14) = 6P (from assumption)

Therefore LHS = 3.4^k + 6P

= 6Q, where Q = 1/2.4^k + P (We take out 6 as a common factor. This works because 4^k is always an even integer, meaning that half of it will also be an integer)

Therefore true for n=k+1 if true for n=k, and since true for n=1, is hence true by mathematical induction for all n≥1.


I_F

 

[wccchick]

sorry still confused - where did the 6P and QP come from?

 

[insert-username]

just a query - where did you get the 6P and the 6Q from?

It's from the "multiple of 6". Basically, a multiple of six is the product of 6 and another integer, and I just used P and Q as that other integer.


I_F

 

[wccchick]

oh i get it!
thanks heaps!