- #1
AN630078
- 242
- 25
- 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?
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?