Induction Help (2 Viewers)

[Green Yoda]

Prove by induction that 12^n>7^n + 5^n for all integers n>=2

 

[Drongoski]

Prove by induction that 12^n>7^n + 5^n for all integers n>=2

You can easily show: true for n = 2

To save time, I'll show only the key steps.

We assume true for n = k >= 2

i.e. 12^k > 7^k + 5^k

.: 12^k+1 > 12 x {7^k + 5^k} = 12 x 7^k + 12 x 5^k
> 7 x 7^k + 5 x 5^k = 7^k+1 + 5^k+1

 

[Green Yoda]

You can easily show: true for n = 2

I am stuck in the s(k+1) part as I cant manipulate it to sub in the the assumed true s(k)

 

[Drongoski]

I am stuck in the s(k+1) part as I cant manipulate it to sub in the the assumed true s(k)

That's why I showed you only the key part: the part that usually presents the problem.

 

[Green Yoda]

I am still quite unclear..

 

[InteGrand]

I am still quite unclear..

With which part? Drongoski's first step was to multiply both sides of the inductive hypothesis by 12.

 

[pikachu975]

Prove by induction that 12^n>7^n + 5^n for all integers n>=2

Arrange to get 12^n - 7^n - 5^n > 0

Prove true for n=2 that's easy

Assume true for n = k
12^k - 7^k - 5^k > 0

Prove true for n = k + 1
LHS = 12^(k+1) - 7^(k+1) - 5^(k+1)
= 12(12^k) - 7(7^k) - 5(5^k)
Factorise the 12 out and pay back extra terms
= 12 (12^k - 7^k - 5^k) + 5(7^k) + 7(5^k)
and since 12^k - 7^k - 5^k > 0 (assumption) and k>0, then
12 (12^k - 7^k - 5^k) + 5(7^k) + 7(5^k) > 0

Therefore proved by mathematical induction