• May 28th 2008, 04:25 AM
acid_rainbow
Hey all this is my first problem i have posted

here is the question

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

a) Prove that n=6k+2

Thanks, i really do not know where to start

thanks
• May 28th 2008, 04:35 AM
CaptainBlack
Quote:

Originally Posted by acid_rainbow
Hey all this is my first problem i have posted

here is the question

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

a) Prove that n=6k+2

Thanks, i really do not know where to start

thanks

Start by writting down the two coefficients in question:

Coeff of $\displaystyle x^2$ : $\displaystyle {n \choose 2}(2k)^{n-2}$

Coeff of $\displaystyle x^3$ : $\displaystyle {n \choose 3}(2k)^{n-3}$

So you have:

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

RonL
• May 28th 2008, 04:38 AM
acid_rainbow
yea i got that far, but how do i solve/ rearrange that to get n=6k+2
• May 28th 2008, 04:45 AM
acid_rainbow
how can you simplify something if its NC2 :SSSSS

so confused
• May 28th 2008, 05:03 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Start by writting down the two coefficients in question:

Coeff of $\displaystyle x^2$ : $\displaystyle {n \choose 2}(2k)^{n-2}$

Coeff of $\displaystyle x^3$ : $\displaystyle {n \choose 3}(2k)^{n-3}$

So you have:

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

RonL

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

so:

$\displaystyle \frac{1}{2!(n-2)!} (2k) = \frac{1}{3!(n-3)!}$

$\displaystyle \frac{1}{(n-2)} (2k) = \frac{1}{3}$

which will rearrange into what you want

RonL
• May 28th 2008, 05:06 AM
acid_rainbow
thanks, sorry for double post, im not sure where the 1 comes from but i will just learn it.