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Math Help - Question relating to the coefficients of the expansion of (2k + x)^n

  1. #1
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    Question relating to the coefficients of the expansion of (2k + x)^n

    Hello,

    The question is:

    In the binomial expansion of (2k + x)^n, where k is a constant and n is a positive integer, the coefficient of  x^2 is equal to the coefficient of  x^3. Prove that n = 6k + 2.

    I got as far as:

    ^{n}C_{2}(2k)^{n - 2} = ^{n}C_{3}(2k)^{n - 3}

    I tried simplifying and got:

    ^{n}C_{2}= ^{n}C_{3}(2k)^{-1}

    I then tried simplifying the factorials, but couldn't see how I was going to end up with the required result.

    Some help would be very appreciated.

    Thank you.
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  2. #2
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    Re: Question relating to the coefficients of the expansion of (2k + x)^n

    Quote Originally Posted by Furyan View Post
    Hello,

    The question is:

    In the binomial expansion of (2k + x)^n, where k is a constant and n is a positive integer, the coefficient of  x^2 is equal to the coefficient of  x^3. Prove that n = 6k + 2.

    I got as far as:

    ^{n}C_{2}(2k)^{n - 2} = ^{n}C_{3}(2k)^{n - 3}

    I tried simplifying and got:

    ^{n}C_{2}= ^{n}C_{3}(2k)^{-1}

    I then tried simplifying the factorials, but couldn't see how I was going to end up with the required result.

    Some help would be very appreciated.

    Thank you.
    Each term in the expansion is of the form \displaystyle \begin{align*} {n\choose{r}} \left( 2k \right)^{n - r} \left( x \right) ^r \end{align*}

    In order for the \displaystyle \begin{align*} x^2 \end{align*} coefficient to be the same as the \displaystyle \begin{align*} x^3 \end{align*} coefficient, that would mean you have to have

    \displaystyle \begin{align*} {n \choose{2} } \left( 2k \right)^{n - 2} &= {n \choose{3} } \left( 2k \right)^{n - 3} \\ \left[ \frac{n!}{2! \left( n - 2 \right)!} \right] \left( 2k \right)^{n - 2} &= \left[ \frac{n!}{3! \left( n - 3 \right)!} \right] \left( 2k \right)^{n - 3} \\ \left[ \frac{n!}{2! \left( n - 2 \right)! } \right] \left[ \frac{\left( 2k \right)^{n-2}}{\left( 2k \right)^{n-3}} \right] &= \frac{n!}{3! \left( n - 3 \right)! } \\ \left[ \frac{n!}{2! \left( n - 2 \right)! } \right] \left( 2k \right) &= \frac{n!}{3! \left( n - 3 \right)!} \\ \frac{2k}{2! (n - 2)! } &= \frac{1}{3! (n - 3)!} \\ \frac{k}{(n - 2)!} &= \frac{1}{6(n - 3)!} \\ k &= \frac{(n - 2)!}{6(n - 3)!} \\ k &= \frac{(n - 2)(n-3)!}{6(n - 3)!} \\ k &= \frac{n - 2}{6} \\ 6k &= n - 2 \\ 6k + 2 &= n \end{align*}
    Thanks from Furyan
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    Member Furyan's Avatar
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    Re: Question relating to the coefficients of the expansion of (2k + x)^n

    Hello Prove It,

    Wow! Thank you so much, that's totally brilliant. I didn't anticipate such a comprehensive reply. I see that by dividing by 2k^{n-3}, instead, you avoided getting (2k)^{-1}. Why didn't I think of that? As for what you did to simplify the factorials, I'm going to have to work through it slowly to make sure I understand what you have done.

    Thank you again. I really appreciate your efforts.
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