Finding the Independent Term in a Binomial Expansion

[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.

 

[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.

 

[dextercioby]

HINT:

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

Answer:k=4...

Daniel.

 

[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

 

[Zurtex]

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]

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:

[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]

~~

 

[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.

 

[xanthym]

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]

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.