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    Hi everyone,

    I'm hoping you can help me with a certain style of C2 Binomial Expansion.

    I would like to know how to solve -

    (a+bx)^p

    When you have values for a & b, and a coefficient of a term in the expansion but you need to find the value of p.

    What would be the procedure of doing this?

    Thanks - Andy.
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    Need more info.
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    (Original post by Circadian_Rhythm)
    Need more info.
    Sorry - I forgot something - you know the coefficient of a term in the expansion also.
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    Well, what would you do if you had P and needed to find the coefficient. Just do the same and you should get an equation you can solve.
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    hey there
    I believe you can do this by expanding it and then equating the two coefficients

    Coefficient of  x^2 in  (1-2x)^p = 40
    where P is greater than 0
    So expanding to find  x^2
    



\dbinom{P}{2}(1)^{P-2}(-2x)^2

    I personally prefer the NCr formula but i can't seem to get it on latex.
    Anyway
     1^{p-2} = 1



So 

\dbinom{P}{2}(-2)^2= 40

 

finding the value of \dbinom{P}{2} = \frac{P!}{(P-2)!2!}



= \dfrac{P(P-1)(P-2)!}{(P-2)!2!}

(P-2)! cancel out

So you are left with 

\dfrac{P(P-1)}{2}

\dfrac{P^2-P}{2}



so going back to \dbinom{P}{2}(-2)^2= 40

We now can get to 

4(\dfrac{P^2-P}{2})=40

    I'm pretty sure you can solve the rest yourself. Good luck!

    For future reference and cause i'm lazy this was Question 4 Exercise 5D of the C2 edexcel book

    And apologies if this counts as a full solution, I aimed to do it as an example to the OP to show the method and did not go all the way to finding the answer
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    The coefficient of which term, it's simply working backwards. Is it the a^2 term? etc?
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    (Original post by MinpoloD)
    hey there
    I believe you can do this by expanding it and then equating the two coefficients

    Coefficient of  x^2 in  (1-2x)^p = 40
    where P is greater than 0
    So expanding to find  x^2
    



\dbinom{P}{2}(1)^{P-2}(-2x)^2

    I personally prefer the NCr formula but i can't seem to get it on latex.
    Anyway
     1^{p-2} = 1



So 

\dbinom{P}{2}(-2)^2= 40

 

finding the value of \dbinom{P}{2} = \frac{P!}{(P-2)!2!}



= \dfrac{P(P-1)(P-2)!}{(P-2)!2!}

(P-2)! cancel out

So you are left with 

\dfrac{P(P-1)}{2}

\dfrac{P^2-P}{2}



so going back to \dbinom{P}{2}(-2)^2= 40

We now can get to 

4(\dfrac{P^2-P}{2})=40

    I'm pretty sure you can solve the rest yourself. Good luck!

    For future reference and cause i'm lazy this was Question 4 Exercise 5D of the C2 edexcel book

    And apologies if this counts as a full solution, I aimed to do it as an example to the OP to show the method and did not go all the way to finding the answer
    I think you made it too complicated when you could have just used n(n-1) /1x2 x^2 formula and from there its easier but i do see ur point
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    (Original post by Education_1)
    I think you made it too complicated when you could have just used n(n-1) /1x2 x^2 formula and from there its easier but i do see ur point
    why reply to a thread which is over 6 months old just to to tell me my method is too complicated?
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    (Original post by MinpoloD)
    why reply to a thread which is over 6 months old just to to tell me my method is too complicated?
    Sorry MinpoloD i didnt realised that this thread was over 6 months (I'm kinda getting used to student room).. no dont minunderstand me I was just saying for the benefit of other people who is seeing this because my teacher didnt actually explain anything .. >< anyway your post of helpful I must admit because I was stuck on some question and I saw your reply.. thank you for explaining .. I didnt mean to offend
 
 
 
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