sryy another binomial q help plss (1 Viewer)

tickboom

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Could you differentiate both sides? And then note that for the RHS, when r=0 that term drops off? I think that gets you there …
 

yashbb

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Could you differentiate both sides? And then note that for the RHS, when r=0 that term drops off? I think that gets you there …
but then what do i do with the r=1, do i start with that as the first term?
 

tickboom

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Could you differentiate both sides? And then note that for the RHS, when r=0 that term drops off? I think that gets you there …
Oh actually I think there’s still an issue with the x^r instead of x^(r-1) on the RHS … hmmmm
 

tickboom

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Unfortunately nothing jumps out ... I did a quick crack at induction but no luck ...
 

tickboom

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Oh wait, I think my first idea does work ... Just needed a few little extra steps ...

Pic.jpg
 

Life'sHard

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Maybe create one thread dedicated to just binomial questions whenever you have a question you can post them there.
 
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CM_Tutor

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Perhaps easier to see without the sigma notation:



I can re-write this theorem as:


Now, we are given that:









A minor quibble with @tickboom's approach, mostly for the sake of MX2 students... saying that , which is effectively the calculation


involves the implicit assumption that , because you can't divide by zero. Thus, Tickboom's proof has actually shown that


and to be completed, needs the case to be addressed separately. Now, that case is trivial in that it has LHS = 0 and RHS = 0, and would quite likely be ignored in an MX1 question... but in an MX2 question under the proof topic, some markers would note that the proof was incomplete. My proof, above, does not have any such problem as it multiplies by rather than dividing. It is true that multiplying by 0 must be done with caution (as it can take a false statement and produce a true one), but in this case, it is taking a statement that is known to be true, in which case multiplying by any (including ) will not produce a false statement, and so the proof covers all real . Note also that my proof could be written more concisely... I have separated the differentiation steps to make what is happening clearer and more explanatory... I would write a briefer proof in an exam situation, something like:



I can re-write this theorem as:



Starting from the given binomial expansion:



An Alternative Proof

Anyone familiar with the standard proof might recognise, or notice from the result on RHS of the theorem, that there is an extra . So , start with:


and, by multiplying by :







We have thus established that:


And thus know that either or we have our required result...
 

yashbb

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Perhaps easier to see without the sigma notation:



I can re-write this theorem as:


Now, we are given that:









A minor quibble with @tickboom's approach, mostly for the sake of MX2 students... saying that , which is effectively the calculation


involves the implicit assumption that , because you can't divide by zero. Thus, Tickboom's proof has actually shown that


and to be completed, needs the case to be addressed separately. Now, that case is trivial in that it has LHS = 0 and RHS = 0, and would quite likely be ignored in an MX1 question... but in an MX2 question under the proof topic, some markers would note that the proof was incomplete. My proof, above, does not have any such problem as it multiplies by rather than dividing. It is true that multiplying by 0 must be done with caution (as it can take a false statement and produce a true one), but in this case, it is taking a statement that is known to be true, in which case multiplying by any (including ) will not produce a false statement, and so the proof covers all real . Note also that my proof could be written more concisely... I have separated the differentiation steps to make what is happening clearer and more explanatory... I would write a briefer proof in an exam situation, something like:



I can re-write this theorem as:



Starting from the given binomial expansion:



An Alternative Proof

Anyone familiar with the standard proof might recognise, or notice from the result on RHS of the theorem, that there is an extra . So , start with:


and, by multiplying by :







We have thus established that:


And thus know that either or we have our required result...
THANK YOUUUU
 

Undefined

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I think that sigma notation is extension 2 now. Nesa writes " Sigma notation is formally introduced in the Mathematics Extension 2 course in the topic MEX-P2: Further Proof by Mathematical Induction. ". You need to still be able to do this question though - I recognize it from a yearly paper I can't remember which though.
 

tickboom

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Whats the purpose of the second line?
The second line differentiates both sides wrt x. The purpose of doing that is to get the exponents to match what we are being asked to prove.
 

=)(=

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The second line differentiates both sides wrt x. The purpose of doing that is to get the exponents to match what we are being asked to prove.
ohh sorry i mean the second last one where you have 0+...
 

tickboom

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ohh sorry i mean the second last one where you have 0+...
ohhhh … that step is to show why you can start the summation from r=1 instead of r=0 (notice the subtle difference in where the sigma starts).
 

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