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Math Help - help: Binomial Theorem - Expanding Terms

  1. #1
    ilc
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    help: Binomial Theorem - Expanding Terms

    The first three terms in the expansion of (1 + ay)^n are 1, 12y and 68y^2.

    Evaluate a and n. Use the fact that:

    (1+ay)^n = 1 + nay + {[n(n-1)]/2}(ay)^2 +....


    ----------------------

    This is as far as I've gotten and I'm not sure what I should do next:

    nay=2y
    => na=12.
    => n=12/a
    => a =12/n

    {[n(n-1)]/2} (ay)^2 = 68 y^2.
    {[n(n-1)]/2} (a)^2 = 68
    {[(n^2) - n]/2} (a)^2 = 68
    [(n^2) - n](a)^2 = 136
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  2. #2
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    Quote Originally Posted by ilc View Post
    The first three terms in the expansion of (1 + ay)^n are 1, 12y and 68y^2.

    Evaluate a and n. Use the fact that:

    (1+ay)^n = 1 + nay + {[n(n-1)]/2}(ay)^2 +....


    ----------------------

    This is as far as I've gotten and I'm not sure what I should do next:

    nay=2y
    => na=12.
    => n=12/a
    => a =12/n

    {[n(n-1)]/2} (ay)^2 = 68 y^2.
    {[n(n-1)]/2} (a)^2 = 68
    {[(n^2) - n]/2} (a)^2 = 68
    [(n^2) - n](a)^2 = 136
    You have 2 unknowns in that.
    If you have only one unknown, you can solve.

    Hence, use n=\frac{12}{a} to write it in terms of "a" only,

    thereby solving for "a".

    Or use a=\frac{12}{n} to write it in terms of "n" only,

    thereby solving for "n".

    Once you've found one, you can find the other.
    This is why the problem gave 2 clues.
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  3. #3
    ilc
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    Quote Originally Posted by Archie Meade View Post
    You have 2 unknowns in that.
    If you have only one unknown, you can solve.

    Hence, use n=\frac{12}{a} to write it in terms of "a" only,

    thereby solving for "a".

    Or use a=\frac{12}{n} to write it in terms of "n" only,

    thereby solving for "n".

    Once you've found one, you can find the other.
    This is why the problem gave 2 clues.

    I understand that I have to use a=12/n or n=12/a and substitute it into the equation to solve it but I'm not sure what to do after I substitute it in.

    a =12/n
    n =12/a

    [(n^2) - n](a)^2 = 136
    [(n^2) - n](12/n)^2 = 136
    what would I do after this?
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  4. #4
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    Quote Originally Posted by ilc View Post
    I understand that I have to use a=12/n or n=12/a and substitute it into the equation to solve it but I'm not sure what to do after I substitute it in.

    a =12/n
    n =12/a

    [(n^2) - n](a)^2 = 136
    [(n^2) - n](12/n)^2 = 136
    what would I do after this?
    Solve for n.
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  5. #5
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    Yes, you must now solve for n.

    I would prioritise knowing how to solve the resulting problem,
    since expanding the binomial expansion is more complex than that.

    understand the following very clearly,
    then return to the beginning of the problem.
    That would be my advice.

    You can also find "a" first...

    an=12\ \Rightarrow\ n=\frac{12}{a}

    \frac{n(n-1)a^2}{2}=68\ \Rightarrow\ \frac{12}{a}\left(\frac{12}{a}-\frac{a}{a}\right)a^2=136

    \frac{12}{a}\left(\frac{12-a}{a}\right)a^2=136

    12(12-a)=136\ \Rightarrow\ 3(12-a)=34\ \Rightarrow\ 12-a=\frac{34}{3}

    a=12-\frac{34}{3}=\frac{36}{3}-\frac{34}{3}=\frac{2}{3}

    Use this "a" to discover "n".
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  6. #6
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    completely lost...

    Could someone explain how to get even this far. I can't even figure out how to START solving this problem.

    "This is as far as I've gotten and I'm not sure what I should do next:

    nay=2y
    => na=12.
    => n=12/a
    => a =12/n"

    Thanks,

    Wyatt...
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