Q3 (c) --> mathematical induction question (1 Viewer)

cssftw · Oct 31, 2011

So i didn't have much experience with inequality induction, I'd like to know what mark/3 I'd get, probs either 1 or 2/3.

$\textup{Prove}\; (2n)!\: \geq \: 2^{n}(n!)^2\\\\\textup{Test for n=1}\\\\\therefore \: \textup{LHS}=(2\times 1)!=2\\\\\textup{RHS}=2^{1}(1!)^2=2\\\\\therefore \textup{LHS}\geq \textup{RHS}\\\\\therefore \textup{True for n=1}\\\\\textup{Assume true for n=k, k integer}\\\\\therefore (2k)!\geq 2^{k}(k!)^2\\\\\textup{Prove for n=k+1}\\\\\therefore (2(k+1))!\\\\=(2k+2)!\\\\=(2k+2)(2k+1)(2k!)\\\\\geq (2k+2)(2k+1)(2^{k}(k!)^2)\\\\=2(k+1)(2k+1)(2^{k}(k!)(k!))\\\\=(2^{k+1})((k+1)!)(2k+1)(k!) ->\textup{It was here I had no idea what to do next - btw note the equal sign instead of the 'greater' sign}\\\\=2^{k+1}[(k+1)!]^2 --> \textup{not too sure if I included this step (might have left it out)}\\\\\therefore \textup{true for n=k+1}\\\\\therefore \textup{true for n=1, true for n=k+1 when true for n=k, k integer}\\\\\therefore \textup{true for n=2, n=3.... true for all positive integers n}$

Obviously I didn't know that 2k+1 > k+1 LOL

So is this 1/3 or 2/3? --> let's say I did forget to put in that final step of $=2^{k+1}[(k+1)!]^2$ and just went on from the previous step straight to the conclusion -- 1/3 or 2/3?

thanks

cutemouse · Oct 31, 2011

2(k+1)! = ... >= (2k+2)(2k+1)2^k (k!)^2

= 2(k+1)(2k+1) 2^k (k!)^2
= (k+1)(2k+1) 2^(k+1) (k!)^2
>= (k+1)(k+1) 2^(k+1) (k!)^2 (as 2k > k for k>=1)
= 2^(k+1) ((k+1)!)^2, as required.

etc

Btw, this is really a 3U induction question...

Q3 (c) --> mathematical induction question (1 Viewer)

cssftw

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cutemouse

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