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Does my logic in this inductionq question make any sense? At all? (1 Viewer)

sinophile

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Prove true for 12^n > 7^n +5^n

After second two steps,

rtp 12^k+1 > 7^k+1 + 5^k+1

From the n=k step,
7^k + 5^k > 7^k+1 + 5^k+1

Therefore true for n=k+1.
Make sense? Bad logic?
 

Drongoski

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Prove true for 12^n > 7^n +5^n

After second two steps,

rtp 12^k+1 > 7^k+1 + 5^k+1

From the n=k step,
7^k + 5^k > 7^k+1 + 5^k+1

Therefore true for n=k+1.
Make sense? Bad logic?
No .. I don't follow. Shouldn't that be: 12^n >= 7^n + 5^n ???

Lets say you you've established for n = 1

Assume true for n = k (k >= 1)
i.e. 12^k >= 7^k + 5^k
therefore:

12^(k+1) = 12 x 12^k >= 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)

Hence shown true for n = k+1 if true for n = k

etc blah blah blah

Ni dau di shi bu shi hua ren ne ?
 

vds700

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Prove true for 12^n > 7^n +5^n

After second two steps,

rtp 12^k+1 > 7^k+1 + 5^k+1

From the n=k step,
7^k + 5^k > 7^k+1 + 5^k+1

Therefore true for n=k+1.
Make sense? Bad logic?
I dont think what u have done is correct.
Step 2. Assume true for n = k
ie 12k > 7k + 5k

Step 3. Prove true for n = k + 1
ie, prove that 12k+1 > 7k+1 + 5k+1
LHS
= 12.12k
>12(7k + 5k)
keep going, u should be able to do it, sorry i cbf right now

EDIT: Ah damn, was beaten to it
 

Drongoski

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vds700 - sorry about that. I appreciate how you feel. As a slow typist it happens to me often. But you have made a useful contribution.

How do you get the superscripts = I'd like to be able to use it for simpler cases so I can avoid LaTeX. Thanks.
 

vds700

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vds700 - sorry about that. I appreciate how you feel. As a slow typist it happens to me often. But you have made a useful contribution.

How do you get the superscripts = I'd like to be able to use it for simpler cases so I can avoid LaTeX. Thanks.
{sup}x{/sup} gives superscript x, by replacing the {} with []
 

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