Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Grazes
Group Title
Find the term in the binomial expansion of (((x^2)/3)(1/x))^9 that is a constant.
 one year ago
 one year ago
Grazes Group Title
Find the term in the binomial expansion of (((x^2)/3)(1/x))^9 that is a constant.
 one year ago
 one year ago

This Question is Closed

Grazes Group TitleBest ResponseYou've already chosen the best response.0
\[(\frac{ x ^{2} }{ 3 }  \frac{ 1 }{ x })^{9}\]
 one year ago

hewsmike Group TitleBest ResponseYou've already chosen the best response.1
Nasty one. The binomial expansion\[(a+b)^{n}= \sum_{k=0}^{n}\frac{ n! }{k!(nk)!}a^{nk}b^{k}\]has powers of a decreasing while those of b increase as you go up in the index k. Consider for this problem that \[a=x^{2}/3\]and \[b= 1/x=x^{1}\]and \[n=9\]They are effectively asking for the value of k that gives\[(a)^{nk}(b)^{k}=(x^{2}/3)^{9k}(x^{1})^{k}=constant \times x^{0}\]and so looking at those powers\[2(9k)+(k)=0\]\[182kk=0\]\[3k=18\]\[k=3\]so what is the multiplier of that power of x?\[(1/3)^{93}(1)^{3}\frac{9!}{3!(93)!}=(1)(3^{6})\frac{(9\times 8 \times 7 \times 6 \times 5\times4 \times 3 \times 2 \times 1)}{(3 \times 2 \times 1)(6 \times 5\times4 \times 3 \times 2 \times 1)}\].... aaargh ......\[=28/3^{5}\]I think ! :)
 one year ago

Grazes Group TitleBest ResponseYou've already chosen the best response.0
I just know that the final answer is 29/8. Thanks!
 one year ago

hewsmike Group TitleBest ResponseYou've already chosen the best response.1
Ah, I've probably fumbled the arithmetic somewhere, but the method is as described. :)
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.