# expansion, binomial

• May 9th 2009, 09:34 AM
Tweety
expansion, binomial
Quote:

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 $\displaystyle x^3$ .

(a) Prove that n=6k+2.
$\displaystyle (2k+x)^n = 2k^n + ^n\mathrm{C}_1 (2k)^{n-1}x +^n\mathrm{C}_2 (2k)^{n-2} x^2 +^n\mathrm{C}_3 (2k)^{n-3}x^3 +....$

Coefficient of $\displaystyle x^2$ = coefficient of $\displaystyle x^3$

$\displaystyle ^n\mathrm{C}_2(2k)^{n-2} = ^n\mathrm{C}_3(2k)^{n-3}$

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

I dont know how to simplify this further to get $\displaystyle n=6k+2$

Any help appreciated,

Thanks.
• May 9th 2009, 09:45 AM
Plato
Quote:

Originally Posted by Tweety
I dont know how to simplify this further to get $\displaystyle 6k=n-2$

$\displaystyle \begin{gathered} 6k = n - 2 \hfill \\ 6k + 2 = n - 2 + 2 \hfill \\ 6k + 2 = n \hfill \\ \end{gathered}$
• May 9th 2009, 09:49 AM
TheEmptySet
Quote:

Originally Posted by Tweety
$\displaystyle (2k+x)^n = 2k^n + ^n\mathrm{C}_1 (2k)^{n-1}x +^n\mathrm{C}_2 (2k)^{n-2} x^2 +^n\mathrm{C}_3 (2k)^{n-3}x^3 +....$

Coefficient of $\displaystyle x^2$ = coefficient of $\displaystyle x^3$

$\displaystyle ^n\mathrm{C}_2(2k)^{n-2} = ^n\mathrm{C}_3(2k)^{n-3}$

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

I dont know how to simplify this further to get $\displaystyle 6k=n-2$

Any help appreciated,

Thanks.

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

$\displaystyle \frac{n!}{ (n-2)!2!} (2k)^{n-2} - \frac{n!}{ (n-3)!3!} (2k)^{n-3} =0$

Now lets factor

$\displaystyle \frac{n!}{ (n-2)!2!} (2k)^{n-2} - \frac{n!}{ (n-3)!3!} (2k)^{n-3} =0 \iff$

$\displaystyle n!(2k)^{n-3}\left( \frac{2k}{(n-2)!2!}-\frac{1}{(n-3)!3!}\right)=0$

Now by the zero factor rule we get

$\displaystyle \left( \frac{2k}{(n-2)!2!}-\frac{1}{(n-3)!3!}\right)=0 \iff \left( \frac{2k}{(n-2)!2!}=\frac{1}{(n-3)!3!}\right)$

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

$\displaystyle 2k=\frac{(n-2)}{3} \iff 6k=n-2$
• May 9th 2009, 09:57 AM
Tweety
^^thanks, Althought I dont know what the 'zero factor rule' is? So I am not really following how you got the desired expression?
• May 9th 2009, 10:04 AM
TheEmptySet
Quote:

Originally Posted by Tweety
^^thanks, Althought I dont know what the 'zero factor rule' is? So I am not really following how you got the desired expression?

You probably know it by a different name

It states that if
$\displaystyle a\cdot b=0$ then $\displaystyle a=0 \mbox{ or } b=0$

It is what you use to solve equations by factoring and setting each factor equal to zero.

We know that exponentials and factorials are never zero so the other factor must be zero.
• May 9th 2009, 10:18 AM
Tweety
Quote:

Originally Posted by TheEmptySet
You probably know it by a different name

It states that if
$\displaystyle a\cdot b=0$ then $\displaystyle a=0 \mbox{ or } b=0$

It is what you use to solve equations by factoring and setting each factor equal to zero.

We know that exponentials and factorials are never zero so the other factor must be zero.

Oh yes I do know that, thanks. Also how comes you are able to factor out $\displaystyle (2k)^{n-3}$ ? Becasue if you multiply it back you wont end up getting $\displaystyle (2k)^{n-2}$
Can you explain ? Thanks!
• May 9th 2009, 10:22 AM
TheEmptySet
Quote:

Originally Posted by Tweety
Oh yes I do know that, thanks. Also how comes you are able to factor out $\displaystyle (2k)^{n-3}$ ? Becasue if you multiply it back you wont end up getting $\displaystyle (2k)^{n-2}$
Can you explain ? Thanks!

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

Remember when we multiply things with the same base we add the exponents