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Coefficients on expansion

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Given that p(x) is a polynomial defined by (x+1)(x^2+2)(x^4+4)(x^8+8)....(x^1024+1024)
    and knowing that the coefficient on x^2012 can be written as 2^a, find a.

    2. Relevant equations

    Binomial thereom mabye idk

    3. The attempt at a solution
    Tried grouping up terms that would sum to 2012 but didnt work nicely.
    The answer is 6 btw
     
  2. jcsd
  3. Feb 7, 2012 #2

    eumyang

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    Expand:
    (x+1)(x2+2) = x3 + x2 + 2x + 2
    (x+1)(x2+2)(x4+4) = ?
    (x+1)(x2+2)(x4+4)(x8+8) = ?

    Do you notice the pattern?
     
  4. Feb 7, 2012 #3
    Yes you would just add all the exponents with each other. I got it but i dont get how that helps me find the coefficient, becuase wouldnt the coefficient be one
     
  5. Feb 7, 2012 #4

    eumyang

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    No, it wouldn't. If you do get the pattern, can you please tell us the answers to my previous post?
    Also, are you sure that the answer is 6? I'm getting a different answer.
     
  6. Feb 7, 2012 #5
    Yes it is 6, it was on question 20 on amc 12 exact question
    mabye you got the pattern wrong the next one is x^16 + 16, x^32+32
    etc

    Yes i get the pattern the first two coefficients will start off as one, as in the highest two, then the next two will have a coefficient of 2, then the next 2 4 all the way to 8 until it is x^0.

    Also recall that it is expressed as 2^a so its 2^6 as a coefficient.
     
  7. Feb 7, 2012 #6

    eumyang

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    Okay, I got the answer of 6. And yes, I did know that the next binomials would be (x16 + 16), (x32 + 32), etc.

    This was why I had a different answer. The pattern is not as obvious as I originally thought.

    Look at the expansion of
    (x+1)(x2+2)(x4+4). (I'll call this P2.)
    This polynomial will have 8 terms. Compare the coefficients of 1st 4 terms with the coefficients of
    (x+1)(x2+2). (I'll call this P1.) What do you notice?
    You will also note that the coefficients of the 2nd 4 terms of P2 are multiples of the coefficients of the 1st 4 terms of P2, respectively.

    Now look at
    (x+1)(x2+2)(x4+4)(x8+8). (I'll call this P3.)
    This polynomial will have 16 terms. Compare the coefficients of 1st 8 terms with the coefficients of P2. What do you notice?
    You will also note that the coefficients of the 2nd 8 terms of P3 are multiples of the coefficients of the 1st 8 terms of P3, respectively.

    I don't know if I can say any more without giving the answer away.
     
  8. Feb 7, 2012 #7
    If you multiply everything in p2 by x^8 you simply get all the terms in P3.
    Also for the first is you multiply it by x^4 you get the first four terms of the next.
     
  9. Feb 7, 2012 #8

    eumyang

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    Note:
    P2 = (x + 1)(x2 + 2)(x4 + 4)

    P3 = (x + 1)(x2 + 2)(x4 + 4)(x8 + 8)
    = P2(x8 + 8)
    = x8(P2) + 8(P2)
    (x8(P2) gives us the 1st 8 terms of P3, and 8(P2) gives us the last 8 terms of P3)

    P4 = (x + 1)(x2 + 2)(x4 + 4)(x8 + 8)(x16 + 16)
    = P3(x16 + 16)
    = x16(P3) + 16(P3)
    (x16(P3) gives us the 1st 16 terms of P4, and 16(P3) gives us the last 16 terms of P4)

    You need to continue the pattern of coefficients until the number of terms in Px exceeds the number of terms between the leading coefficient of P10 and the term that contains x2012.
     
  10. Feb 7, 2012 #9
    I see, but would another valid solution be listing the powers of 2

    so x^1 x^2 x^4 x^8 x^16 x^32 x^64,x^128 x^256 x^512 x^1024

    Then you see if you add 1024 512 256 128 64 16 8 and 4 it sums to 2012, you are left with x^1 x^2 and x^32

    So the number of ways would be 1*2*32
    so 64 or 2^6 which gets 6.

    Is that valid?
     
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