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Math Help - Series expansions

  1. #1
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    Series expansions

    I'm attempting this question on binomial distributions:

     1 + mx + \frac{m(m-1)x^2}{2} are the first 3 terms of the series expansion of (1+x)^m

    Use this to find the 1st 3 terms of the series expansion of  (1+x)^{m+1} (1-2x)^m

    This is what I've done:

    1) Substituted (m+1) into the initial expansion to give the expansion for  (1+x)^{m+1} :

     1 + (m+1)x + \frac{m(m+1)x^2}{2}

    2) Used the binomial theorum to expand  (1 - 2x)^m :

     1 - 2mx + {2m(m-1)x^2}

    3) Is this along the right lines...if so where do I go from here? Or am I doing completely the wrong thing from the start?
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  2. #2
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    Quote Originally Posted by orangeiv View Post
    I'm attempting this question on binomial distributions:

     1 + mx + \frac{m(m-1)x^2}{2} are the first 3 terms of the series expansion of (1+x)^m

    Use this to find the 1st 3 terms of the series expansion of  (1+x)^{m+1} (1-2x)^m

    This is what I've done:

    1) Substituted (m+1) into the initial expansion to give the expansion for  (1+x)^{m+1} :

     1 + (m+1)x + \frac{m(m+1)x^2}{2}

    2) Used the binomial theorum to expand  (1 - 2x)^m :

     1 - 2mx + {2m(m-1)x^2}

    3) Is this along the right lines...if so where do I go from here? Or am I doing completely the wrong thing from the start?
    HI

    I think you are correct . So now multiply the 2 expansions you got .
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    HI

    I think you are correct . So now multiply the 2 expansions you got .
    Ok...so we need the  x^0 , x^1 , and  x^2 coefficients...it's horrendous. This is what I got:

     1 - x(2m-1) + 0.5(m^2 - m)x^2
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  4. #4
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    Quote Originally Posted by orangeiv View Post
    Ok...so we need the  x^0 , x^1 , and x^2 coefficients...it's horrendous. This is what I got:

     1 - x(2m-1) + 0.5(m^2 - m)x^2
    I got :

    [1+(m+1)x+\frac{1}{2}m(m+1)x^2+...][1-2mx+2m(m-1)x^2+...]

    =1-2mx+2m(m-1)x^2+(m+1)x-2m(m+1)x^2+\frac{1}{2}m(m+1)x^2+...

    =1+(m+1-2m)x+[2m(m-1)-2m(m+1)+\frac{1}{2}m(m+1)x^2+...]

    =1+(1-m)x+\frac{1}{2}(m^2-7m)x^2+...
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