Binomial Expansion Help

D

dxs

New member
Joined
Jul 3, 2020
Messages
23
So I was finishing up an assignment and I've been stuck on the last question for about an hour. I think I understand the concept behind this question but the y is throwing me off. Please help
1614374917919.png
 
dxs said:
So I was finishing up an assignment and I've been stuck on the last question for about an hour. I think I understand the concept behind this question but the y is throwing me off. Please help
View attachment 25379
Click to expand...
Please show one attempt, so we can have an idea where you might be going astray, and don't have to guess.
 
I get to
1614375740575.png
where n = 20

and then from there I'm unsure how to solve for r with the y there.
 
The problem is not good English so I cannot be sure what you mean.

But do the substitutions

[MATH]u = x^3 \text { and } v = \dfrac{y^2}{x}[/MATH].

[MATH]\therefore \left ( x^3 + \dfrac{y^2}{x} \right )^{20} = (u + v)^{20} \text { and } x^{36} = WHAT?[/MATH]
 
I presume you chose not to show the binomial coefficient. (Oh, I see. You were just asked for which term, not what the term is.)

Look at the expression you wrote. Simplify it; what power of x do you get? Set that exponent equal to 36, and solve for r.

Another approach you could take would be to factor 1/x out of the base, and take that into account when expanding. But what you are doing is probably more natural. I'd say it probably isn't the presence of y, so much as the presence of x in both terms, that is confusing you.
 
  • Like
Reactions: dxs
Dr.Peterson said:
I presume you chose not to show the binomial coefficient. (Oh, I see. You were just asked for which term, not what the term is.)

Look at the expression you wrote. Simplify it; what power of x do you get? Set that exponent equal to 36, and solve for r.

Another approach you could take would be to factor 1/x out of the base, and take that into account when expanding. But what you are doing is probably more natural. I'd say it probably isn't the presence of y, so much as the presence of x in both terms, that is confusing you.
Click to expand...
Oh crap I forgot about the binomial coefficient. The coefficient would be 20Cr,

and for the x simplified I got 1614379650621.png
 
Subhotosh Khan said:
Remember there is a term containing 'x' in the denominator. Then you would get:

\(\displaystyle x^{60-4r}\)
Click to expand...
So I managed to solve it through with n = 20 and r = 6 and got the x^36y^12, but going back to the original question of "what term number", how would I figure out the term number?
 
You say that the y is bothering so then you are lucky since in this case you can simply ignore the y. Think about it, the y factor does not change the power of x at all so it can be ignored.
 
What does “term number” even mean? What EXACTLY do the problem’s instructions say.
 
dxs said:
So I managed to solve it through with n = 20 and r = 6 and got the x^36y^12, but going back to the original question of "what term number", how would I figure out the term number?
Click to expand...
I've been assuming that [MATH]r[/MATH] is what you are calling the term number. But you'll have to look at the wording of your problem, and the definition of terms in your book to determine what they are asking for.
 
You must log in or register to reply here.
Top