Induction Help (1 Viewer)

Green Yoda

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

Drongoski

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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. 12k > 7k + 5k

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

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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.
 

pikachu975

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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
 

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