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

Hello,

The question is:

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

I got as far as:

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

I tried simplifying and got:

$\displaystyle ^{n}C_{2}= $$\displaystyle ^{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.

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

Quote:

Originally Posted by

**Furyan** Hello,

The question is:

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

I got as far as:

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

I tried simplifying and got:

$\displaystyle ^{n}C_{2}= $$\displaystyle ^{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 $\displaystyle \displaystyle \begin{align*} {n\choose{r}} \left( 2k \right)^{n - r} \left( x \right) ^r \end{align*}$

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

$\displaystyle \displaystyle \begin{align*} {n \choose{2} } \left( 2k \right)^{n - 2} &= {n \choose{3} } \left( 2k \right)^{n - 3} \\ \left[ \frac{n!}{2! \left( n - 2 \right)!} \right] \left( 2k \right)^{n - 2} &= \left[ \frac{n!}{3! \left( n - 3 \right)!} \right] \left( 2k \right)^{n - 3} \\ \left[ \frac{n!}{2! \left( n - 2 \right)! } \right] \left[ \frac{\left( 2k \right)^{n-2}}{\left( 2k \right)^{n-3}} \right] &= \frac{n!}{3! \left( n - 3 \right)! } \\ \left[ \frac{n!}{2! \left( n - 2 \right)! } \right] \left( 2k \right) &= \frac{n!}{3! \left( n - 3 \right)!} \\ \frac{2k}{2! (n - 2)! } &= \frac{1}{3! (n - 3)!} \\ \frac{k}{(n - 2)!} &= \frac{1}{6(n - 3)!} \\ k &= \frac{(n - 2)!}{6(n - 3)!} \\ k &= \frac{(n - 2)(n-3)!}{6(n - 3)!} \\ k &= \frac{n - 2}{6} \\ 6k &= n - 2 \\ 6k + 2 &= n \end{align*}$

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 $\displaystyle 2k^{n-3}$, instead, you avoided getting $\displaystyle (2k)^{-1}$. Why didn't I think of that?(Wondering) 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. (Bow)(Party)