Binomial Expansion: Coefficient of x^3 in (2/x-3x^4)^12

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    Binomial Expansion
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Homework Help Overview

The discussion revolves around finding the coefficient of x^3 in the binomial expansion of (2/x - 3x^4)^12. Participants explore the application of the binomial theorem in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formula for binomial expansion and how to determine the appropriate value of k for the term corresponding to x^3. There are attempts to derive k through manipulation of exponents, and some participants express uncertainty about the inclusion of the binomial coefficient.

Discussion Status

Several participants are actively engaging with the problem, questioning each other's reasoning and clarifying the role of the binomial coefficient. There is a recognition of the need to ensure correct interpretations of the expressions involved.

Contextual Notes

One participant raises a related question about a different expression, leading to confusion regarding the expected terms and their powers. This indicates a broader exploration of binomial expansions and their implications.

TheRedDevil18
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Homework Statement



Find the coefficient of x^3 in the binomial expansion of
(2/x - 3x^4)^12

Homework Equations

The Attempt at a Solution



Expanding this out would take too long and I cannot use a calculator to find the coefficient

I know the formula for the expansion

summation (12 choose k) a^k * b^12-k

a = 2/x, b = -3x^4

But how do I find k ?
 
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For which k does the term correspond to a x^3 term if you insert a and b into your expression?
 
well if I ignore the coefficients, I get

x^-k * x^(48-4k) = x^3
48-5k = 3
k = 9 ?

so my coefficient would be 2^9 (-3)^3 ?
 
Almost, you dropped the binomial coefficient which should also be there.
 
You seem to have forgotten the "binomial coefficient", \begin{pmatrix}12 \\ 9 \end{pmatrix}.
 
Ok, so with the binomial coefficient

(12 choose 9) 2^9 (-3)^3 ?
 
Okay guys, I have another question relating to the same topic

Given (3x - 2/x^3)^40, Find coefficient x^10

I'll skip the plugging into formula for a and b, but here's how I solve for k

x^k * x^-3(40-k) = x^(-120+4k) = x^10
-120+4k = 10
k = 65/2

Now in the memo, they have
-120+4k = -20.....How did they get -20 ?
k = 100/4
 
TheRedDevil18 said:
How did they get -20 ?

This is a very good question ... Just from looking at it for 5 seconds, I do not see the possibility of having a term x^10. Any term should be x^40 multiplied by some power of x^-4 which gives terms x^12 and x^8, but no term x^10.
 
TheRedDevil18 said:

Homework Statement



Find the coefficient of x^3 in the binomial expansion of
(2/x - 3x^4)^12

Homework Equations

The Attempt at a Solution



Expanding this out would take too long and I cannot use a calculator to find the coefficient

I know the formula for the expansion

summation (12 choose k) a^k * b^12-k

a = 2/x, b = -3x^4

But how do I find k ?
You wrote
\left( \frac{2}{x} - 3 x^4 \right)^{12}
Is that what you meant, or did you want
\left( \frac{2}{x - 3 x^4} \right)^{12}?
If the latter, use parentheses, like this: (2/(x - 3x^4))^12 or [2/(x - 3x^4)]^12.
 
  • #10
Ray Vickson said:
You wrote
\left( \frac{2}{x} - 3 x^4 \right)^{12}
Is that what you meant, or did you want
\left( \frac{2}{x - 3 x^4} \right)^{12}?
If the latter, use parentheses, like this: (2/(x - 3x^4))^12 or [2/(x - 3x^4)]^12.

It's the first one
 
  • #11
Orodruin said:
This is a very good question ... Just from looking at it for 5 seconds, I do not see the possibility of having a term x^10. Any term should be x^40 multiplied by some power of x^-4 which gives terms x^12 and x^8, but no term x^10.

So is the question wrong or something ?, I'm just not sure where the -20 came from
 
  • #12
Ok, the question was wrong, it was x^-20. All fine now, thanks guys :)
 

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