# Binomial Theorem question

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.

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Zurtex
Homework Helper
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.

dextercioby
Homework Helper
HINT:

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

Daniel.

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

Zurtex
Homework Helper
tutor69 said:
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
That's a simpler equation ?

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

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

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

Daniel.

xanthym
.

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

Code:
[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:
dextercioby
Homework Helper
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.

xanthym
dextercioby said:
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.
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.

~~

dextercioby