Please show one attempt, so we can have an idea where you might be going astray, and don't have to guess.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
Oh crap I forgot about the binomial coefficient. The coefficient would be 20Cr,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.
Remember there is a term containing 'x' in the denominator. Then you would get:Oh crap I forgot about the binomial coefficient. The coefficient would be 20Cr,
and for the x simplified I got View attachment 25384
Ohh okay and when I set the exponent of 60-4r to 36 I got that r = 6Remember there is a term containing 'x' in the denominator. Then you would get:
\(\displaystyle x^{60-4r}\)
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?Remember there is a term containing 'x' in the denominator. Then you would get:
\(\displaystyle x^{60-4r}\)
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.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?