Harder binomial (1 Viewer)

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From 2004 Reddam House trial - last question. (Solution cut out from file)

.

I have differentiated and subbed x=1...hmm wrong idea??

Thanks
 
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Timske

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Can you post the question here
 
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Yeah - I was just fixing up the latex - it's up now.
 
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RHS = (1/2) ( 2^(2n) + (2n)!/(n!)^2 )

it looks like you might have to sub x=-1 at some point, then add two expressions together.
 

RealiseNothing

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I was thinking that since it is the sum of a combinatoric, then maybe you could somehow use the sum of the rows of Pascal's triangle ( where n is the row).
 

Sy123

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I THINK I have a solution here. Ill edit it in

Here it is, hopefully you will be able to see what I did, and also I dont know how to enter the values for a summation in latex.


 
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Fus Ro Dah

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From 2004 Reddam House trial - last question. (Solution cut out from file)

.

I have differentiated and subbed x=1...hmm wrong idea??

Thanks
This question is taken directly from the Cambridge 3U Year 12 book Exercise 5F Question 17 Page 207.
 
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Actually...Just revisited this.

I think there is a flaw to your solution???

In line 2 of your solution you have but the question is - ie. missing another n! in the denominator...

I'm not sure if this ruins your solution but I may be missing it...The rest looks fine...but unsure.

Thanks!
 
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Sy123

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Very very sorry about that, but its a typo, because


So yeah that line of working was just restating the question, which I did wrong :/

But my solution is still right
 

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