Vandomo's question at Yahoo Answers regarding the binomial theorem

[MarkFL]

Here is the question:

Binomial Expansion - Can Someone Please Help?

In the binomial expansion ( 2K + X) ^n, where K is an constant and n is a +ve integer, the coefficient of X^ 2 is = to the coefficient of X ^3.

Prove that n = 6 K + 2

If someone could please help, I'd be grateful.

Thanks.
Vandomo

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Vandomo,

According to the binomial theorem, we may state:

\(\displaystyle (2k+x)^n=\sum_{j=0}^n\left[{n \choose j}(2k)^{n-j}x^j \right]\)

Now, if the coefficients of the terms containing $x^2$ and $x^3$ are equal, then this implies:

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

Divide through by \(\displaystyle (2k)^{n-3}\) to get:

\(\displaystyle {n \choose 2}(2k)={n \choose 3}\)

Use the definition \(\displaystyle {n \choose r}\equiv\frac{n!}{r!(n-r)!}\) to now write:

\(\displaystyle \frac{n(n-1)}{2}(2k)=\frac{n(n-1)(n-2)}{6}\)

Multiply through by \(\displaystyle \frac{6}{n(n-1)}\) to obtain:

\(\displaystyle 6k=n-2\)

Arrange as:

\(\displaystyle n=6k+2\)

Shown as desired.