92 3U HSC Q6c)ii): Binomial Question⌄ (1 Viewer)

gamja

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1992 3U HSC Q6c)ii)
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from , I integrated both sides to get


and then divided both sides by x to get


and subbed in x=-1 to get LHS correct, but RHS became zero...

Could someone tell me where I went wrong? Thanks in advance!! :D
 

pikachu975

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So when you integrated you need the +C, so you have:

x + (1/2)(nC1)x^2 + ... = (1+x)^(n+1)/(n+1) + C
Sub in x = 0
C = -1/(n+1)

I'm not sure if you can divide both sides by x as x could be 0, but you could multiply both sides by x:

x^2 + (1/2)(nC1)x^3 + ... = x/(n+1) (1+x)^(n+1) - x/(n+1) = x/(n+1) ((1+x)^(n+1) - 1)

Now subbing x = -1:

LHS is correct still, RHS = -1/(n+1) (0 - 1) = 1/(n+1)
 

5uckerberg

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1992 3U HSC Q6c)ii)
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from , I integrated both sides to get


and then divided both sides by x to get


and subbed in x=-1 to get LHS correct, but RHS became zero...

Could someone tell me where I went wrong? Thanks in advance!! :D
I have a different take on this one, given the pattern of the binomial question you will have

Integrate that and you will have


Here,



Divide by x and we will have


Which is equivalent to


Next step sub in

Then you have
 

hogzillaAnson

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This is probably the most 'condensed' version of what you could do, but in an exam I wouldn't count on being able to see it. The above methods of subbing in to find C are much more reliable.



As to why we can interchange summation and integration, this follows from what's called the 'additivity of the definite integral'. For example,



So you can easily imagine that



In other words,



Later on, one will see that there are important restrictions on what's being summed to make this true, but as far as Maths Extension 1 (and 2) go, they never really arise.
 

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