Equation Simplification Help Please...

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Equation Simplification Help Please...

Mathematics
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\[(2(k+1)!/4^{k+1})\times(4^{k}/(2k)!)\]
4^k/4^(k+1) simplifies to 1/4, which means that your expression simplifies to (k+1)!/(2(2k)!).
Sorry! Just realised I missed out a pair of brackets!\[((2(k+1))!/4^{k+1})\times(4^{k}/(2k)!)\]

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Other answers:

is it: \[(2(k+1))!/4(2k)!\]
\[[2(k+1)]! = [2k+2]*[2k+1]*[2k]!\] See if that doesn't help simplify.
I think I get (4k^2+5k+2)/4
Close. Should be \[(2k+1)(k+1)/2 = (2k^2 + 3k + 1)/2\]
How did you get that? Aren't we left with the [2k+2]∗[2k+1] on top
yes, and 4 on bottom. You can divide 2k+2 by 2, leaving (k+1)*(2k+1) on top and 2 on the bottom.
Haha! What's wrong with me today... That was obvious! Thanks for all your help mate. Any chance you could send me your email or something so I can contact you when I don't find you here? You've been a massive help to me today!
http://dpaste.com/526875/
Let me know when you get it. and send me an email.
got it mate.
Send me a quick email and I'll delete the dpaste.
Sent.

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