krypticlemonjuice
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In this question, how did they get from that line to the next. This is for part (iii)
Integration Q Howard (1 Viewer)
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In this question, how did they get from that line to the next. This is for part (iii)
They subbed in the necessary values to make it (2n-1)! and (2n-2)! by multiplying top and bottom (ie multiplying (2n-2)(2n-4)...2 top and bottom)
then they just write In as I0 where n is 0
and then they solve
the multiplying top and bottom is a very common trick
krypticlemonjuice
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Hmm yeah that would be great! I get what you explained. Now i dont' really understand the last four lines of working out in that q
I'll do it nowwww
Look I'm ngl I think their solution is a bit wrong, it doesn't read right...
krypticlemonjuice
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Oh damn, would you be able to show me a correct way??
In the arrow part they're just converting it to a double factorial because every second number is present. The step after that is a bit trickier but it makes sense if you break it into the numerator and denominator.
!! \text{ is all of the odd numbers up to 2n} \\\therefore (2n-1)!! =\frac{\text{all numbers up to 2n}}{\text{all even numbers up to 2n}} \\ =\frac{(2n)!}{2^n(n!)} \text{ this works because we can effectively seperate each term of }2^nn!\text{ into }2\times 1\text{, }2\times 2\text{ and so on until }2\times n\text{ which is all of the even numbers until 2n}\\\text{By this same principle, given that }(2n-2)!!\text{ is just all of the even numbers until }2n-2 \text{, we can just divide by the same thing again for all even numbers} \\\therefore (2n-1)!! =\frac{\frac{(2n)!}{2^n(n!)}}{2^n(n!)} )
The rest is trivial
The rest is trivial