Binomial Expansion: Evaluating Coefficient from two binomials

[AN630078]

Homework Statement

Hello, I have been practising binomial expansion problems, specifically related to finding the coefficient of values of x. I came across the problem below and found that I rather confused myself in solving it. Could anyone evaluate my method to suggest if there are any improvements to be made here? I would be very grateful

Find the coefficient of x^3 in the expansion of (3-5x)(1+1/3x)^18

Relevant Equations

nCr

So I think I may be overcomplicating this problem but I realize that in order to find the x^3 term it will be the product of the two binomials, ie. x^1*x^2=x^3. The coefficient of x^3 will be the coefficient of x^1 in the first bracket multiplied by the coefficient of x^2 in the second bracket.
Since the first binomial is to the power of 1 we can assume the value of the x term if the second binomial is x^2.
So the next step would be finding when this occurs in (1+1/3)^18:
18C2* (1)^16*(1/3x)^2=153*1*1/9x^2
Then multiply this by the first bracket:
(3-5x)(153*1*1/9x^2)=-255 x^3

So the coefficient of x^3 would be -255.

Would this be correct?

 

[Master1022]

You are forgetting about the contribution of the $x^3$ term in the expansion of $(1 + \frac{1}{3} x)^{18}$ which will be multiplied with the 3 from the preceding binomial

In your first binomial $(3 - 5x)$ you have $3x^0$ and $-5x^1$. You are looking for the final coefficient of $x^3$ so you need to look for both the $x^2$ and $x^3$ terms in $(1 + \frac{1}{3} x)^{18}$.

Hope that makes some sense. If not, then I am happy to clarify.

[EDIT]: Apologies, I typed the same binomial out twice, have corrected it now. However, the method still stands as suggested.

 

[AN630078]

You are forgetting about the contribution of the $x^3$ term in the expansion of $(1 + \frac{1}{3} x)^{18}$ which will be multiplied with the 3 from the preceding binomial

In your first binomial $(3 - 5x)$ you have $3x^0$ and $-5x^1$. You are looking for the final coefficient of $x^3$ so you need to look for both the $x^2$ and $x^3$ terms in $(1 + \frac{1}{3} x)^{18}$.

Hope that makes some sense. If not, then I am happy to clarify.

[EDIT]: Apologies, I typed the same binomial out twice, have corrected it now. However, the method still stands as suggested.

Thank you for your reply. Sorry I am a little confused, could you elaborate further?

 

[Master1022]

Thank you for your reply. Sorry I am a little confused, could you elaborate further?

Sure, so when you expand $(1 + \frac{1}{3} x)^{18}$ we will get something like $a_0 + a_1 x^1 + a_2 x^2 + a_3 x^3 + ...$. When we multiply that by $(3 - 5x)$, the $-5x$ will combine with the $a_2 x^2$ term to make $-5 a_2 x^3$ and the $3$ will combine with $a_3 x^3$ to form $3 a_3 x^3$. Thus the resulting coefficient of $x^3$ is $(-5 a_2 + 3 a_3 )$. That means we will need to consider the coefficients of the $x^2$ and $x^3$ terms in the expansion of $(1 + \frac{1}{3} x)^{18}$

 

[AN630078]

Sure, so when you expand $(1 + \frac{1}{3} x)^{18}$ we will get something like $a_0 + a_1 x^1 + a_2 x^2 + a_3 x^3 + ...$. When we multiply that by $(3 - 5x)$, the $-5x$ will combine with the $a_2 x^2$ term to make $-5 a_2 x^3$ and the $3$ will combine with $a_3 x^3$ to form $3 a_3 x^3$. Thus the resulting coefficient of $x^3$ is $(-5 a_2 + 3 a_3 )$. That means we will need to consider the coefficients of the $x^2$ and $x^3$ terms in the expansion of $(1 + \frac{1}{3} x)^{18}$

Thank you for reply. Admittedly, I am still a little confused but far less than I was. I think I understand what you mean though.
So I should use the binomial formula to expand (1+1/3)^18=1+6x+17x^2+272x^3/9+340x64/9...
Then multiply this by the first bracket:
(3-5x)(1+6x+17x^2+272x^3/9+...)
Which if you expand the brackets is equal to;
3(1+6x+17x^2+272x^3/9+...)-5x(1+6x+17x^2+272x^3/9+...)
3+18x+51x^2+272x^3/3-5x-30x^2-85x^3-1360x^4/9
3+13x+21x^2+17x^3/3-1360x^4/9

So would the coefficient of x^3 would be 17/3?

 

[epenguin]

You didn't need to write out all those terms.
As already explained by Master 1022 who has almost answered the problem question for you, there are only two terms in the expansion of the second bracket that are going to give you a x³ term after multiplication by the first bracket.

(Hope you have copied out the original question complete with its brackets correctly.)

 

[AN630078]

You didn't need to write out all those terms.
As already explained by Master 1022 who has almost answered the problem question for you, there are only two terms in the expansion of the second bracket that are going to give you a x³ term after multiplication by the first bracket.

(Hope you have copied out the original question complete with its brackets correctly.)

Thank you for your. Oh, so I only need to gather the like terms of x^3;

(3*272x^3/9)+(-5x*17x^2)=17x^3/3
Therefore, the coefficient of x^3 is 17/3

 

[epenguin]

Yes I get the same result.