A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

rational
 10 months ago
Best ResponseYou've already chosen the best response.0n = 1 : \(a_1 = 1 \lt 2^1 \) so the given statement is true for n = 1

rational
 10 months ago
Best ResponseYou've already chosen the best response.0ty :) im clueless on next step :(

rational
 10 months ago
Best ResponseYou've already chosen the best response.0would it be like induction proof from first principle : assume \(a_k = a_{k1} + a_{k2} + a_{k3}\lt 2^k\) and prove \(a_{k+1} \lt 2^k\)

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2I would do three cases for the base case. For a1,a2,and a3. You want to assume your statement is true for every integer out to some chosen integer value k and then you want to show it is true for k+1. So let's consider the case of k+1 \[a_{k+1}=a_k+a_{k1}+a_{k2}\] we need to look at a_(k1) and a_k So we know a_k<2^k and a_(k1)<2^(k1) by our assumption

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2oops and a_(k2) so we have also a_(k2)<2^(k2)

rational
 10 months ago
Best ResponseYou've already chosen the best response.0oh that makes sense 3 base cases ! and yes using second principle we have : \(a_{k} \lt 2^{k} \) \(a_{k1} \lt 2^{k1}\) \(a_{k2} \lt 2^{k2} \)

rational
 10 months ago
Best ResponseYou've already chosen the best response.0\(a_{k+1}=a_k+a_{k1}+a_{k2} \lt 2^k + 2^{k1} + 2^{k2}\)

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2yep good so far do you want to try to figure out the trick into showing <2^(k+1)?

rational
 10 months ago
Best ResponseYou've already chosen the best response.0I'm trying at the moment lol, need to make the right hand side 2^{k+1}

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2I used what I wanted to show to help me start out I don't know if that helps you

rational
 10 months ago
Best ResponseYou've already chosen the best response.0im getting right hand side as : \(2^{k2}(2^2 + 2 + 1) = 2^{k2}(7)\)

rational
 10 months ago
Best ResponseYou've already chosen the best response.0not sure how to conclude... only if that 7 was 8... :o

rational
 10 months ago
Best ResponseYou've already chosen the best response.0it has to be \(\ge 8 \) i think

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2I liked factoring out 2^(k+1) instead of 2^(k2) but you should be able to proceed as you are

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2and nip is right we want <

nipunmalhotra93
 10 months ago
Best ResponseYou've already chosen the best response.1you have your term <2^(k2) (7)<2^(k2) (8)=2^(k+1)

rational
 10 months ago
Best ResponseYou've already chosen the best response.0so far i have this : \(a_{k+1}=a_k+a_{k1}+a_{k2} \lt 2^k + 2^{k1} + 2^{k2} = 2^{k2}(7)\)

nipunmalhotra93
 10 months ago
Best ResponseYou've already chosen the best response.1yeah factoring out 2^(k+1) will do too.

rational
 10 months ago
Best ResponseYou've already chosen the best response.0thanks a lot that was easier than it looked initally to start wid lol xD

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2or I liked: \[a_{k+1}<2^k+2^{k1}+2^{k2}=2^{k+1}(2^{1}+2^{2}+2^{3}) \] \[=2^{k+1}(\frac{1}{2}+\frac{1}{4}+\frac{1}{8})=2^{k+1}(\frac{4+2+1}{8})=2^{k+1}\frac{7}{8}<2^{k+1} \frac{8}{8}=2^{k+1}(1)=2^{k+1}\] Both ways are cute

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2both ways require you to know 7<8 that is interesting

rational
 10 months ago
Best ResponseYou've already chosen the best response.0hahah yes this looks easy to read quick :) thank you both :D

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2hey rational here is another proof you can look at that has the same style we just worked with http://php.scripts.psu.edu/djh300/cmpsc360/exstronginduction.htm

nipunmalhotra93
 10 months ago
Best ResponseYou've already chosen the best response.1you're welcome (Y)

rational
 10 months ago
Best ResponseYou've already chosen the best response.0the proof in that link is very good !! cleared up the mechanics of strong induction thank you very much @myininaya
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.