# Binomial Theorem Question

• September 10th 2008, 10:07 AM
ecopetition
Binomial Theorem Question
Quote:

Prove that n has no solutions given that r=2 when:
$
n!/
r!(n-r)!
= 4
$

I expanded this to:
$
n(n-1)(n-2)(n-3)...1/
2((n-2)(n-3)(n-4)...1)
= 4
$

Which cancels down to:
$
n(n-1)/
2
= 4
$

Which goes to:

$
n^2-n
= 8
$

Which goes to:

$
n^2-n - 8
= 0
$

This is now a quadratic which, when treated with the discriminant (a=1, b=-1, c=-8)
$b^2-4ac$
$(-1)^2-4*1*-8$
$=1+32 = 33$

So, this quadratic has real solutions, but the questions asks to prove it does not. What have I done wrong?

Thanks.
• September 10th 2008, 10:46 AM
Laurent
Try to finish writing down the solutions of your quadratic equation, and check if it makes sense to put them back in your initial problem with factorials...
• September 10th 2008, 10:51 AM
Laurent
Afterwards, have a look at the wikipedia about the link between factorials and the "Euler Gamma function" (if you don't know it). This explains what sense could anyway be given to your solutions (the positive one, and even the negative one).
• September 10th 2008, 11:09 AM
ecopetition
Ok, my answers are:

$1/2(1-\sqrt(33))$ and
$1/2(1+\sqrt(33))$

Both give 4 when whacked into the original equation.

I remember a teacher saying that $n$ must be an integer? But this would appear to be untrue according to Google's calculator?
• September 10th 2008, 11:25 AM
Shyam
Quote:

Originally Posted by ecopetition
I expanded this to:
$
n(n-1)(n-2)(n-3)...1/
2((n-2)(n-3)(n-4)...1)
= 4
$

Which cancels down to:
$
n(n-1)/
2
= 4
$

Which goes to:

$
n^2-n
= 8
$

Which goes to:

$
n^2-n - 8
= 0
$

This is now a quadratic which, when treated with the discriminant (a=1, b=-1, c=-8)
$b^2-4ac$
$(-1)^2-4*1*-8$
$=1+32 = 33$

So, this quadratic has real solutions, but the questions asks to prove it does not. What have I done wrong?

Thanks.

See here,

after completely solving we get,
$
n= \frac{1+\sqrt{33}}{2}, \frac{1-\sqrt{33}}{2}$

n= 3.37, - 2.37

we know that factorial $n!$ can be calculated only for positive integers only, so n should be a positive integer.
Since n is not a positive integer, So the given equation does not have any solution.
• September 10th 2008, 01:05 PM
Laurent
Quote:

Originally Posted by ecopetition
I remember a teacher saying that $n$ must be an integer? But this would appear to be untrue according to Google's calculator?

$n!=n(n-1)\cdots 2\cdot 1$ is the product of the first $n$ positive integers. So the usual definition makes no sense if $n$ is not an integer and your teacher is fully right.

How would one anyway generalize it for any real number in a "nice way"? Two answers:

- in fact, you are considering a binomial coefficient ${n\choose k}$ and these still make sense if $n$ is any real number, since: ${n\choose k}=\frac{n(n-1)\cdots(n-k+1)}{k!}$ (the number of terms is $k$, which is an integer).

- refering to my last post, you can use Euler's Gamma function to extend the factorial, like Google does. This is a function defined on $(0,+\infty)$ by $\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt$ (well, it can even by defined for other real or even complex numbers, but let's keep simple). Using integration by part, one can show that for every $n\in\mathbb{N}$, $\Gamma(n+1)=n!$. So this function is a candidate for a generalization of the factorial. In fact, for every $x>0$, $\Gamma(x+1)=x\Gamma(x)$, and $\Gamma$ is a smooth, convex (even log-convex) function. As a conclusion, under various requirements, $\Gamma$ appears to be the most natural extension of the factorial: $x!=\Gamma(x+1)$. But usually we write it with $\Gamma$ and not with the exclamation mark "!".