Finding k for the coefficient of x^3 in (2/x−3x^4)^12

[TheRedDevil18]

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 ?

 

[Orodruin]

For which k does the term correspond to a x^3 term if you insert a and b into your expression?

 

[TheRedDevil18]

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 ?

 

[Orodruin]

Almost, you dropped the binomial coefficient which should also be there.

 

[HallsofIvy]

You seem to have forgotten the "binomial coefficient", $$\begin{pmatrix}12 \\ 9 \end{pmatrix}$$.

 

[TheRedDevil18]

Ok, so with the binomial coefficient

(12 choose 9) 2^9 (-3)^3 ?

 

[TheRedDevil18]

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

 

[Orodruin]

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.

 

[Ray Vickson]

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.

 

[TheRedDevil18]

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

 

[TheRedDevil18]

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

 

[TheRedDevil18]

Ok, the question was wrong, it was x^-20. All fine now, thanks guys :)