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

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

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

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

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

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

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

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

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

myininaya
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1you're welcome (Y)

rational
 2 years 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
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