A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
use the binomial theorem to show that
2^n > n(n1)(n2)/3!
true for all n in natural numbers
anonymous
 5 years ago
use the binomial theorem to show that 2^n > n(n1)(n2)/3! true for all n in natural numbers

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Couple of tricks here. Let's take the right hand side \[\frac{n(n1)(n2)}{3!}\] Multiply both numerator and denominator by (n3) \[\frac{n(n1)(n2)(n3)!}{(n3)!3!}\] The numerator clearly becomes n! Thus, the RHS can be written as : \[{n}\choose3\] Now, write 2^n as \[(1+1)^n\] If you expand it, it will look like: \[{n \choose 0} 1^n + {n \choose 1}1^{(n1)}1^1 + {n \choose 2}1^{n2}1^2 + {n \choose 3} 1^{n3}1^3 + ...\] The fourth term in this expansion is basically the RHS. So, by this manipulation, we find that 2^n = RHS + some positive terms (cause n is natural), hence we prove the given inequality.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:O thats like so easy.........haha thanks i totally missed that answer on my exam then
Ask your own question
Sign UpFind more explanations on OpenStudy
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.