AEA 2014 paper question 4 (further maths question)

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StopFindingMe
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Came across this question on an AEA paper.
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However I got a bit stuck.

When I checked the mark scheme, I had the same sort of answer, but I don't know how they got from their second line of working out to their third.

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Any insight would be appreciated. Cheers
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RDKGames
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(Original post by StopFindingMe)
Came across this question on an AEA paper.

However I got a bit stuck.

When I checked the mark scheme, I had the same sort of answer, but I don't know how they got from their second line of working out to their third.


Any insight would be appreciated. Cheers
In the entire product, (-1)^r occurs twice so they cancel out to give the (1).

Without any manipulation, you also have r! in the denominator as well as 2.2.2\ldots 2 = 2^r. Hence you got a r!2^r in the denominator.

The manipulation now comes in by rewriting

1.3.5\ldots(2r-1) = \dfrac{1.2.3.4.5\ldots(2r-1)(2r)}{2.4\ldots(2r)}.

Now, carefully, you should note that this denominator is indeed 2^r r!.

This, along with r!2^r from above, makes up the overall new denominator of r! 2^r \times 2^r r!.
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StopFindingMe
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(Original post by RDKGames)
In the entire product, (-1)^r occurs twice so they cancel out to give the (1).

Without any manipulation, you also have r! in the denominator as well as 2.2.2\ldots 2 = 2^r. Hence you got a r!2^r in the denominator.

The manipulation now comes in by rewriting

1.3.5\ldots(2r-1) = \dfrac{1.2.3.4.5\ldots(2r-1)(2r)}{2.4\ldots(2r)}.

Now, carefully, you should note that this denominator is indeed 2^r r!.

This, along with r!2^r from above, makes up the overall new denominator of r! 2^r \times 2^r r!.
Wow, I think I understand, gonna digest it a bit more by trying it again.

How did you manage to have that intuition so quickly? Have you done this question before?
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RDKGames
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(Original post by StopFindingMe)
Wow, I think I understand, gonna digest it a bit more by trying it again.

How did you manage to have that intuition so quickly? Have you done this question before?
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