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Thread: Binomial Expansion

  1. #1 Binomial Expansion 
    Forum Bachelors Degree x(x-y)'s Avatar
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    Hello, I'm currently trying to do a question and have become a bit stuck!

    Find the binomial expansion of:



    I have got this so far, and I'm not sure if it is correct and where to go from here:



    So, I'm not too sure whether this is right- and if it is right, I don't know where to lead on from this. Help, but not the answer, would be appreciated!

    Thanks in advance.


    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
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  3. #2  
    Forum Junior c186282's Avatar
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    Things look good:


    where:


    and in your case and

    The binomial coefficients come from something called Pascal's triangle

    For fun see if you can show


    One can even make a 3D generalization of a Pascal's triangle.
    This is something I made using Asymptote
    3D Pascal's triangle

    One can even make an nD version they just get harder to draw.


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  4. #3  
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    Hmmm... Kind of makes sense to me, but I haven't learn it like that (perhaps we will next year- I'm currently doing AS Level Mathematics).

    We just based the binomial expansion off of Pascal's Triangle- which in the case of my question of (3x-2)^4, the coefficients would be 1 4 6 4 1, as far as I know?

    And so, we just have to find the expansion of the brackets rather than the sum of the expansion- which is what you seem to be showing, I may be wrong.

    By the way, what is 'j'? I've never seen that when doing binomial expansions.

    We've also learnt that:



    But I'm not sure if that would help in this case?

    Thanks for the help so far anyway, it's much appreciated!
    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
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  5. #4  
    Forum Junior c186282's Avatar
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    Going off of you new avatar picture I'm going to guess that you have done some programing.
    Code:
    sum=0;
    for (i=1;i<10;i++){
    sum=sum+i;
    }
    print sum
    The letter "i" could very easily be replaced by any other letter. In math we would call it a dummy variable.
    The mathematical equivalent to the code above is:

    Where I have arbitrarily changed the dummy variable name to j

    So the j in the sum above is just a dummy variable.

    Have you seen factorial?

    Edit thanks to Heinsbergrelatz factional factorial
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  6. #5  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    Have you seen factional
    Slight Typo i am guessing, it is factorial not factional.
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  7. #6  
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    Ah, yes, ok- indeed I have used sigma notation and summations, I just went a bit brain-dead about the variable "j" then! And, also, yes I have indeed used factorials (mostly during statistics).

    Unfortunately, though, I haven't done any programming- I just liked the avatar, sorry!

    Thanks for the help by the way!
    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
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  8. #7  
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    Oh yes, I got a bit confused on another question I was doing too:

    A sequence of terms



    is defined by



    1) Write down the values of

    I did this part and got the following answers:

    .

    Which, to me, seem correct.

    Then the question asks:

    2) Find



    Here I tried using the formula:



    But, the thing I can't work out is what d (the difference) is- so I got a bit stuck...
    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
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  9. #8  
    Forum Junior c186282's Avatar
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    Forget about the formula and look at your pattern and I bet you can write down the answer.
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  10. #9  
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    I can see that it is continuously alternating between 2 and -1, however I don't exactly see how that would help me find the sum of the first 100 terms altogether.

    Wait, a minute- isn't the answer 50?
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  11. #10  
    Forum Junior c186282's Avatar
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    Thats what I got
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  12. #11  
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    Yeah, thought so. Something just clicked and I realised how to get it! Thanks anyway!
    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
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  13. #12  
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    I got the answer to my first question in the OP, by the way- and I'm pretty sure I got it correct.

    My Answer:

    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
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  14. #13  
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    Just one more question I got stuck on whilst doing a past paper, sorry!

    Given that



    show that



    where the values of the constants a and b are to be found.

    I shall tell you the formulae that I know, which may come in useful here:









    Now I'm pretty sure that the above last law of logarithms will be required.

    So, how I'd start the proof would be:



    However, I'm afraid to say that I don't know where to lead on from here! :?

    It may just be a case of "braindeadedness" again, and the proof may be glaringly obvious to me in a while!
    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
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  15. #14  
    Forum Junior c186282's Avatar
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    Lets back up one step

    Now apply

    then


    Only the 10 was raised to the power


    Your expansion looks good.
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  16. #15  
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    ------------------




    "Mathematicians stand on each other's shoulders."- Carl Friedrich Gauss


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  17. #16  
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    Quote Originally Posted by Heinsbergrelatz




    Yes, I obtained that answer recently- looks like I got it right- I did it a slightly different method though! Thanks anyway!
    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
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  18. #17 Bunomial expansion 
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    Please I have problem expanding (1-2x)^(1+x/2)^5
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  19. #18  
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    Please I've got problem expanding (1-2x)^(1+x/2)^5 I'll be pleased to have an immediate reply thanks.
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  20. #19  
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    Quote Originally Posted by tyrex
    Please I've got problem expanding (1-2x)^(1+x/2)^5 I'll be pleased to have an immediate reply thanks.
    I think your are missing something

    What is ?
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  21. #20  
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    sorry n is 3
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  22. #21  
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    My first response has enough information for you to work out your answer.
    However you need to expand each element by itself first, (Do first then the other one)
    Then multiply the final two expansions.

    Some hints that you already know:


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  23. #22  
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    pls can you simplify both together for me cos i still have problem solving it
    (1-2x)^3(1+x/2)^5
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  24. #23  
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    I will do the first term and you do the next then we will talk about the final step.

    Let :
    and









    Below is the code for what I have shown above. Just add the "/" to the trailing [tex] block to make it work.

    [tex]a=1[tex] and [tex] b=-2x[tex]

    [tex](1-2 x)^{3}=\sum_{m=0}^{3}\binom{3}{m} (1)^m(-2x)^{3-m}[tex]

    [tex](1-2 x)^{3}=\binom{3}{0} (1)^{0}(-2x)^{3}+\binom{3}{1} (1)^{1}(-2x)^{2}[tex][tex]+\binom{3}{2} (1)^{2}(-2x)+\binom{3}{3} (1)^{3}(-2x)^{0}[tex]

    [tex](1-2 x)^{3}=-8 \frac{3!}{0!\times 3!}x^3+4 \frac{3!}{1!\times 2!}x^2-2\frac{3!}{2!\times 1!}x+\frac{3!}{3!\times 0!}[tex]

    [tex](1-2 x)^{3}=-8 x^3+12 x^2-6x+1[tex]
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  25. #24  
    Forum Radioactive Isotope MagiMaster's Avatar
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    You should probably put the powers on the 1s to make it a bit clearer what's going on.
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  26. #25  
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    Quote Originally Posted by MagiMaster
    You should probably put the powers on the 1s to make it a bit clearer what's going on.
    Good point, I made the edit above.
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