Finding the Independent Term in a Binomial Expansion

In summary: 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:
  • #1
laaah
3
0
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
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
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
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
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
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
.

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
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
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
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.
 

1. What is the Binomial Theorem?

The Binomial Theorem is a mathematical formula used to expand binomials, which are expressions with two terms, raised to a certain power. It provides a systematic way to find the coefficients of each term in the expansion.

2. What is the general form of the Binomial Theorem?

The general form of the Binomial Theorem is (a + b)^n = a^n + na^(n-1)b + (n(n-1)/2!)a^(n-2)b^2 + ... + b^n, where a and b are the two terms, and n is the power to which the binomial is raised.

3. How is the Binomial Theorem useful in mathematics?

The Binomial Theorem is useful in various areas of mathematics, such as algebra, calculus, and probability. It allows us to easily expand binomial expressions and solve problems involving binomial coefficients.

4. Can the Binomial Theorem be applied to more than two terms?

No, the Binomial Theorem only applies to binomials, which have two terms. However, there are other similar methods, such as the Multinomial Theorem, that can be used for expressions with three or more terms.

5. How can the Binomial Theorem help in simplifying complicated expressions?

The Binomial Theorem can help simplify complicated expressions by providing a systematic way to expand and collect like terms. This allows us to easily identify patterns and simplify the expression to a more manageable form.

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