Recurrence formula (1 Viewer)

[asianese]

It is a common question, I've seen it 2 times in papers...I can't seem to get the last part. I know that I need to times top/bottom by other terms but not sure. Any help is appreciated. Don't worry about latex, a method is just as good =D Thanks







I've got i) easily, but the 2nd part..hmm. I've expanded and tried a couple of things but didn't work.

 

[RealiseNothing]

I got it, posting up the solution now.

 

[asianese]

Thanks a trillion!!!

 

[Alkenes]

he didnt post shit :S

Thanks a trillion!!!

 

[asianese]

He's gonna, I know

 

[RealiseNothing]

Thus:



If we keep doing this, we will eventually get:



Now evaluate

It becomes the integral of 1 since anything to the power of 0 is 1. So the primitive is just 'x', and hence by subbing in limits

So now we have, if we sub in what value got for



Let's take out the for now.

The whole thing becomes:



Factor out all the 2's from the denominator, we will have 'n' amount of them:





The denominator now has n!:



Now we notice that the numerator is just

We now sub this in for the numerator, realise how the the denominator of the above is the same as the denominator we have so far, and so we square the denominator?



So when we square the denominator it becomes:



Put back in the



As required.

 

[asianese]

oh WOW! Thanks so much!! If this comes up in my test ... SHEBANG! Thanks!!

 

[RealiseNothing]

Btw, where did you get this question from?

 

[Alkenes]

WOW!! too much effort

Thus:



If we keep doing this, we will eventually get:



Now evaluate

It becomes the integral of 1 since anything to the power of 0 is 1. So the primitive is just 'x', and hence by subbing in limits

So now we have, if we sub in what value got for



Let's take out the for now.

The whole thing becomes:



Factor out all the 2's from the denominator, we will have 'n' amount of them:





The denominator now has n!:



Now we notice that the numerator is just

We now sub this in for the numerator, realise how the the denominator of the above is the same as the denominator we have so far, and so we square the denominator?



So when we square the denominator it becomes:



Put back in the



As required.

 

[Drongoski]

Excellent work Realise - esp the laborious LaTeX part. Unable to rep you again - had to spread . . . .

 

[zhiying]

Think I saw this in HSC like umm 2006? Can't remember exactly

 

[Siddy123]

Thus:



If we keep doing this, we will eventually get:



Now evaluate

It becomes the integral of 1 since anything to the power of 0 is 1. So the primitive is just 'x', and hence by subbing in limits

So now we have, if we sub in what value got for



Let's take out the for now.

The whole thing becomes:



Factor out all the 2's from the denominator, we will have 'n' amount of them:





The denominator now has n!:



Now we notice that the numerator is just

We now sub this in for the numerator, realise how the the denominator of the above is the same as the denominator we have so far, and so we square the denominator?



So when we square the denominator it becomes:



Put back in the



As required.

good job man!!!!!!!

unfortunately...You must spread some Reputation around before giving it to RealiseNothing again.