Binomial Theorem question

  • Thread starter laaah
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  • #1
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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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  • #2
Zurtex
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Well the long way to do is expand the full thing. But there are easier ways to do it, for example rewrite it as:

[tex]\left(x^2 - 2x^{-1}\right)^6[/tex]

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:

[tex]a\left(x^2\right)^6 + b\left(x^2\right)^5 x^{-1} + \ldots[/tex]

Simplifying:

[tex]ax^{12} + bx^{9} + \ldots[/tex]

Carry on like that until you get a x0 term and work out what the coefficient is.
 
  • #3
dextercioby
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HINT:

[tex] (a-b)^{n}=\sum_{k=0}^{n} C_{n}^{k} (-1)^{k}b^{k}a^{n-k} [/tex]

Answer:k=4...

Daniel.
 
  • #4
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simplify the equation :
f(x) = h(x).g(x)

[tex](\frac{2}{x})^6 ({1- \frac{x^{3}}{2})^6[/tex]

Use the binomial expansion for g(x) and simplify the solution
 
  • #5
Zurtex
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tutor69 said:
simplify the equation :
f(x) = h(x).g(x)

[tex](\frac{2}{x})^6 ({1- \frac{x^{3}}{2})^6[/tex]

Use the binomial expansion for g(x) and simplify the solution
That's a simpler equation :confused: ?
 
  • #6
dextercioby
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My formula delivers the result in 2 lines,on of which is

[tex] -k+12-2k=0 [/tex]

Then computing [itex] C_{6}^{4} [/itex] is elementary.

Daniel.
 
  • #7
xanthym
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.

And for those who would rather not compute [tex] \mathbb{C}_{4}^{6}} [/tex], 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:
  • #8
dextercioby
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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
xanthym
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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 [tex] \mathbb{C}_{4}^{6} = (6!)/[(4!)(2!)] [/tex] ; or
2) Those who want ALL the coefficients in 10 sec; or
3) Those who compulsively doodle.


~~
 
  • #10
dextercioby
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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...:wink:

Daniel.
 

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