Proof by Induction for Inequalities Watch

Y12_FurtherMaths
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Can someone check if my proof for question 9 is valid please?
https://imgur.com/a/nnSuIuq
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Y12_FurtherMaths
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mqb2766
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I didn't fully follow the second column for 9, but at the bottom of the first column it looks like you could do a similar trick for k^2 > k+1, and get straight to the answer without expanding? Other than that it looks ok.
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RDKGames
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(Original post by Y12_FurtherMaths)
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It's valid, but you should really justify why it is true that 3k^3 + 5k^2 + 2k > 3k^2 + 8k + 5
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Y12_FurtherMaths
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(Original post by RDKGames)
It's valid, but you should really justify why it is true that 3k^3 + 5k^2 + 2k > 3k^2 + 8k + 5
Would I do that by saying 3k^3>3k^2, 5k^2>8k, 2k>5 therefore all of it together is true?
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RDKGames
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(Original post by Y12_FurtherMaths)
Would I do that by saying 3k^3>3k^2, 5k^2>8k, 2k>5 therefore all of it together is true?
Yes, keeping in mind that k > 4 makes these hold. I think the first and last are obvious enough, but you could be a bit more explicit as to show that the middle one is true as well. But maybe that's just me being pedantic.

Though I must say, you very much depended on working backwards from the result to the expanded form in order to see what you need to bound your current expression by. This is OK but I think a more natural way to the result would be the following:

$\begin{align*} (k+1)! = (k+1)k! & > (k+1)(3k^2+2k) \\ & = 3(k+1)k^2 + 2(k+1)k \\ & > 3(k+1)(k+1) + 2(k+1) \\ & = 3(k+1)^2 + 2(k+1)\end{align*}$

where the penultimate line is true because of two things:

- (first term) k^2 > k^2 - 1 = (k+1)(k-1) > k+1 therefore k^2 > k+1

- (second term) k > 4 > 1 therefore k > 1.
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Y12_FurtherMaths
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(Original post by RDKGames)
Yes, keeping in mind that k > 4 makes these hold. I think the first and last are obvious enough, but you could be a bit more explicit as to show that the middle one is true as well. But maybe that's just me being pedantic.

Though I must say, you very much depended on working backwards from the result to the expanded form in order to see what you need to bound your current expression by. This is OK but I think a more natural way to the result would be the following:

$\begin{align*} (k+1)! = (k+1)k! & > (k+1)(3k^2+2k) \\ & = 3(k+1)k^2 + 2(k+1)k \\ & > 3(k+1)(k+1) + 2(k+1) \\ & = 3(k+1)^2 + 2(k+1)\end{align*}$

where the penultimate line is true because of two things:

- (first term) k^2 > k^2 - 1 = (k+1)(k-1) > k+1 therefore k^2 > k+1

- (second term) k > 4 > 1 therefore k > 1.
Ok I appreciate this thank you
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