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

1. ## 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 = 6k + 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.

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

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 = 6k + 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*}

3. ## 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? 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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# (2k x)^n ,where k is the constant,the coeficient of x^2 is equal to the coeficient of x^3.prove that n=6k 2

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