Help with an equation please (1 Viewer)

[Elise8842]

Hi,

Could someone please tell me how's the LHS simplified to the RHS? Thanks in advance


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[PhysicsMaths]

Hi,

Could someone please tell me how's the LHS simplified to the RHS? Thanks in advance


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Here's my go at it:
http://imgur.com/rlh0DfW

Not too sure (I've never heard of a double factorial until today)
Step 3 might not work

But basically what I did was split (2n)(2n-1)(2n-2)(2n-3)(2n-4)....3.2.1
into two components; [(2n)(2n-2)(2n-4)(2n-6)...][(2n-1)(2n-3)...] (hence isolating the (2n-1)!!
but the 2^n was a bit tricky

 

[faisalabdul16]

Hi,

Could someone please tell me how's the LHS simplified to the RHS? Thanks in advance


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Okay firstly the double exclamation means this! <img src="http://latex.codecogs.com/gif.latex?(2n-1)!!&space;=&space;\\&space;(2n-1)(2n-3)(2n-5)...&space;etc" title="(2n-1)!! = \\ (2n-1)(2n-3)(2n-5)... etc" />

Now - <img src="http://latex.codecogs.com/gif.latex?(2n)!&space;=&space;\\&space;(2n)(2n-1)(2n-2)(2n-3)(2n-4)...&space;\&space;etc" title="(2n)! = \\ (2n)(2n-1)(2n-2)(2n-3)(2n-4)... \ etc" />

You see how in some of the brackets you can take out 2? You're going to have 2x2x2x2x all the way up to n which is the same as saying <img src="http://latex.codecogs.com/gif.latex?2^{n}" title="2^{n}" />

<img src="http://latex.codecogs.com/gif.latex?\frac{(n!)(2^{n})}{(n!)((2n-1)!!)2^{n}}&space;\\\\&space;=&space;RHS" title="\frac{(n!)(2^{n})}{(n!)((2n-1)!!)2^{n}} \\\\ = RHS" />

Cancel out and you'll get RHS

 

[Elise8842]

Okay firstly the double exclamation means this! <img src="http://latex.codecogs.com/gif.latex?(2n-1)!!&space;=&space;\\&space;(2n-1)(2n-3)(2n-5)...&space;etc" title="(2n-1)!! = \\ (2n-1)(2n-3)(2n-5)... etc" />

Now - <img src="http://latex.codecogs.com/gif.latex?(2n)!&space;=&space;\\&space;(2n)(2n-1)(2n-2)(2n-3)(2n-4)...&space;\&space;etc" title="(2n)! = \\ (2n)(2n-1)(2n-2)(2n-3)(2n-4)... \ etc" />

You see how in some of the brackets you can take out 2? You're going to have 2x2x2x2x all the way up to n which is the same as saying <img src="http://latex.codecogs.com/gif.latex?2^{n}" title="2^{n}" />

<img src="http://latex.codecogs.com/gif.latex?\frac{(n!)(2^{n})}{(n!)((2n-1)!!)2^{n}}&space;\\\\&space;=&space;RHS" title="\frac{(n!)(2^{n})}{(n!)((2n-1)!!)2^{n}} \\\\ = RHS" />

Cancel out and you'll get RHS

Here's my go at it:
http://imgur.com/rlh0DfW

Not too sure (I've never heard of a double factorial until today)
Step 3 might not work

But basically what I did was split (2n)(2n-1)(2n-2)(2n-3)(2n-4)....3.2.1
into two components; [(2n)(2n-2)(2n-4)(2n-6)...][(2n-1)(2n-3)...] (hence isolating the (2n-1)!!
but the 2^n was a bit tricky

Thank you very much!!


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