Mathematical Induction Problem. (1 Viewer)

[CHUDYMASTER]

Show that 12^n > 7^n + 5^n where n >= 2

 

[kini mini]

There is probably a more elegant way to do this problem, but this is the first that occured to me.

(I'll skip the intro and so on that Billy Pender loves so much )

...let k be a positive integer for which the result holds, we prove the result for n = k + 1

To prove: 12^(k+1) > 7^(k+1) + 5^(k+1)

(given that 12^k > 7^k + 5^k)

LHS = 12^(k+1)
= 12.12^k

RHS = 7^(k+1) + 5^(k+1)
= 7.7^k + 5.5^k

now LHS > 12.7^k + 12.5^k from the assumption

and 12.7^k + 12.5^k > RHS

so LHS > RHS and everyone lived happily ever after.

I realise I've skipped a lot of steps, but that's the essence of it...I hope .

 

[CHUDYMASTER]

now LHS > 12.7^k + 12.5^k from the assumption

and 12.7^k + 12.5^k > RHS

so LHS > RHS and everyone lived happily ever after.

Erm...no, you've lost me on those 3 lines...

 

[Ultimate]

Originally posted by CHUDYMASTER
Show that 12^n > 7^n + 5^n where n >= 2

(1) Prove for n = 2,

LHS = 12^2 = 144
RHS = 7^2 + 5^2 = 74

thus LHS>RHS for n = 2.

(2) Assume true for n = k

12^k > 7^k + 5^k

(3) Prove True for n = k+1
i.e. 12^(k + 1) > 7^(k+1) + 5^(k+1)

LHS = 12^k. 12
= 12 (7^k + 5^k) ......from (2) since 12^k > 7^k + 5^k
= 12 . 7^k + 12 . 5^k

RHS = 7. 7^k + 5. 5^k

clearly 12 . 7^k + 12 . 5^k > 7. 7^k + 5 . 5^k

hence LHS > RHS for n = k + 1

(4) If true for n = 2, then the statement is also true for n = k, n = k + 1 and al interger values for n >= 2

 

[kini mini]

Originally posted by CHUDYMASTER


Erm...no, you've lost me on those 3 lines...

Sorry mate, I think I skipped a bit too quickly. Ultimate has written out the solution in full for you now (thanks!).

 

[CHUDYMASTER]

Ahh I get it. Thanks both of you!