Binomial Theorem question

1. Mar 28, 2005

laaah

Hi, I'm having some problems with this question -

Find the term independent of x in

(x^2 - 2/x)^6

I know the answer is something like 6Csomething 2^something, but I'm not sure how to get that. So far I've only really done simple things like (x+y)^n where y is an integer and not another x term. Help would be appreciated.

2. Mar 28, 2005

Zurtex

Well the long way to do is expand the full thing. But there are easier ways to do it, for example rewrite it as:

$$\left(x^2 - 2x^{-1}\right)^6$$

Expand that without looking at the coefficients (and I inculde the sign of the term when I say coefficient) and you get the general expansion:

$$a\left(x^2\right)^6 + b\left(x^2\right)^5 x^{-1} + \ldots$$

Simplifying:

$$ax^{12} + bx^{9} + \ldots$$

Carry on like that until you get a x0 term and work out what the coefficient is.

3. Mar 28, 2005

dextercioby

HINT:

$$(a-b)^{n}=\sum_{k=0}^{n} C_{n}^{k} (-1)^{k}b^{k}a^{n-k}$$

Daniel.

4. Mar 28, 2005

tutor69

simplify the equation :
f(x) = h(x).g(x)

$$(\frac{2}{x})^6 ({1- \frac{x^{3}}{2})^6$$

Use the binomial expansion for g(x) and simplify the solution

5. Mar 28, 2005

Zurtex

That's a simpler equation ?

6. Mar 28, 2005

dextercioby

My formula delivers the result in 2 lines,on of which is

$$-k+12-2k=0$$

Then computing $C_{6}^{4}$ is elementary.

Daniel.

7. Mar 28, 2005

xanthym

.

And for those who would rather not compute $$\mathbb{C}_{4}^{6}}$$, there's always Pascal's Triangle:

Code (Text):

[B]
1
1   1
1   2   1
1    3   3    1
1    4    6    4   1
1   5    10   10   5   1
[COLOR=Red]1   6   15   20   15   6   1[/COLOR]
[/B]

~~

Last edited: Mar 28, 2005
8. Mar 28, 2005

dextercioby

Rather than drawing the Pascal triangle,don't u think it's easier to multiply 6 by 5 and divide the result by 2...?

Daniel.

9. Mar 28, 2005

xanthym

For n<10, Pascal's Triangle is easily derived in 10 sec and serves the needs of:
1) Those who would take longer than 10 sec to remember that $$\mathbb{C}_{4}^{6} = (6!)/[(4!)(2!)]$$ ; or
2) Those who want ALL the coefficients in 10 sec; or
3) Those who compulsively doodle.

~~

10. Mar 28, 2005

dextercioby

I could multiply & divide natural #-s without using a paper.Imagining Pascal's triangle & doing the addition and having the overview is harder,for me,at least...

Daniel.