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Math Help - Binomial expansion help

  1. #1
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    Wink Binomial expansion help

    i have two questions i need help with.
    for the first one should i use pascals triangle?

    1. Expand, fully, each of the following
    i) (x + 2y)^4
    2. Show that, if x is small enough for x^2 and higher powers of x to be neglected, the function (x – 2)(1 + 3x)^8 has a linear approximation of - 2 – 47x.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by perryman View Post
    i have two questions i need help with.
    for the first one should i use pascals triangle?

    1. Expand, fully, each of the following
    i) (x + 2y)^4
    2. Show that, if x is small enough for x^2 and higher powers of x to be neglected, the function (x – 2)(1 + 3x)^8 has a linear approximation of - 2 – 47x.
    If you like, I find it easier to use factorials

    (a+b)^n = {n \choose 0}a^nb^0 + {n \choose 1}a^{n-1}b + {n \choose 2}a^{n-2}b^2 + {n \choose 3}a^{n-3}b^3 + ... + {n \choose n}a^0b^n

    Of course {n \choose 0} = {n \choose n} = b^0 = a^0 = 1
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  3. #3
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    Smile

    generally...
    (x+y)^n=\sum_{k=0}^{n}\begin{pmatrix}<br />
n\\ <br />
k<br />
\end{pmatrix}x^{n-k}y^{k}
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  4. #4
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    for the first question i got the answer x^4 + (8x^3 * y) + (24x^2 *y^2) +32xy^3 + 16y^4 can anyone confirm this
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  5. #5
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    Hello perryman
    Quote Originally Posted by perryman View Post
    i have two questions i need help with.
    for the first one should i use pascals triangle?

    1. Expand, fully, each of the following
    i) (x + 2y)^4
    2. Show that, if x is small enough for x^2 and higher powers of x to be neglected, the function (x – 2)(1 + 3x)^8 has a linear approximation of - 2 – 47x.
    e^(i*pi) and Raoh have given you the formula you need to expand #1. Do you understand how to use it?

    To give you an example, and to help you with #2, I'll expand (1+3x)^8, ignoring powers of x above x^2, using the formula the others have given you:
    (1+3x)^8 = \binom801^8(3x)^0 + \binom811^7(3x)^1 +\binom821^6(3x)^2 + ...
    Looks scary, doesn't it? But remember that 1^n = 1 for all values of n,\; (3x)^0 = 1 and
    \binom80 = 1

    \binom81=8


    \binom82 = \frac{8.7}{2!}

    So, it's simply:
    1 +8(3x) + \frac{8.7}{2!}(3x)^2 + ...
    =1+24x+... (and I've left the x^2 term out now, because we don't need it anyway - I just put it there as an example)
    So if we ignore x^2 and higher powers of x:
    (x-2)(1+3x)^8 = (x-2)(1+24 x + ...)
    Can you complete it now?

    Grandad

    PS I've just read your answer to #1. Good work!
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