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    Given that the expansion (1+ax)^n begins with 1+36x+576x^2 find the values of a and n
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    (Original post by osayuki)
    Given that the expansion (1+ax)^n begins with 1+36x+576x^2 find the values of a and n
    Can you expand (1+ax)^n in terms of a and n?

    It starts (1+ax)^n = 1 + nax + ...

    Then compare the coefficients.


    E.g. the second term in the expansion is 36x so na = 36.

    Does that make sense?
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    (Original post by notnek)
    Can you expand (1+ax)^n in terms of a and n?

    It starts (1+ax)^n = 1 + nax + ...

    Then compare the coefficients.


    E.g. the second term in the expansion is 36x so na = 36.

    Does that make sense?
    Thank you for replying. I'm unsure how to get nax as I couldn't figure out how to cancel n!/(n-1!)1!
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    (Original post by osayuki)
    Thank you for replying. I'm unsure how to get nax as I couldn't figure out how to cancel n!/(n-1!)1!
     \frac{n!}{(n-1)!} (We can "ignore" the 1! since it just equals 1)

    Remembering what factorial actually means and writing n! and (n-1)! out fully we get;

     \frac{n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1}{(n-1) \times (n-2) \times ... \times 3 \times 2 \times 1}

    Cancelling terms which occur on both lines are left with;

     n
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    (Original post by DylanJ42)
     \frac{n!}{(n-1)!} (We can "ignore" the 1! since it just equals 1)

    Remembering what factorial actually means and writing n! and (n-1)! out fully we get;

     \frac{n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1}{(n-1) \times (n-2) \times ... \times 3 \times 2 \times 1}

    Cancelling terms which occur on both lines are left with;

     n
    Thank you so much
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    (Original post by osayuki)
    Thank you for replying. I'm unsure how to get nax as I couldn't figure out how to cancel n!/(n-1!)1!
    It will make your life easier if you remember the pattern of the coefficients so you don't have to derive each one.

    E.g. for the simple expansion (1+x)^n, the second term coefficient is 'n', the third is n(n-1)/2!, the fourth is n(n-1)(n-2)/3! etc.

    This may be on your formula book.
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    (Original post by notnek;[url="tel:62561237")
    62561237[/url]]It will make your life easier if you remember the pattern of the coefficients so you don't have to derive each one.

    E.g. for the simple expansion (1+x)^n, the second term coefficient is 'n', the third is n(n-1)/2!, the fourth is n(n-1)(n-2)/3! etc.

    This may be on your formula book.
    Thank you again for your help
 
 
 
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