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

Ilovecake Group TitleBest ResponseYou've already chosen the best response.0
correct:)
 6 months ago

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

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

myininaya Group TitleBest ResponseYou've already chosen the best response.2
I 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
 6 months ago

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

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

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

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

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

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

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

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

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
use 7<8
 6 months ago

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

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

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

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

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

rational Group TitleBest ResponseYou've already chosen the best response.0
Ohhh yess !
 6 months ago

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

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

myininaya Group TitleBest ResponseYou've already chosen the best response.2
or 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
 6 months ago

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

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
yeah. :D
 6 months ago

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

myininaya Group TitleBest ResponseYou've already chosen the best response.2
hey 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
 6 months ago

nipunmalhotra93 Group TitleBest ResponseYou've already chosen the best response.1
you're welcome (Y)
 6 months ago

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