Hello,
The question is:
In the binomial expansion of, where k is a constant and n is a positive integer, the coefficient of
is equal to the coefficient of
. Prove that n = 6k + 2.
I got as far as:
^{n - 2} = )
I tried simplifying and got:
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 formOriginally Posted by Furyan
Hello,
The question is:
In the binomial expansion of
I got as far as:
I tried simplifying and got:
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.
In order for thecoefficient to be the same as the
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*}](http://latex.codecogs.com/png.latex?\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*})
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, instead, you avoided getting
. 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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