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n = 1 : \(a_1 = 1 \lt 2^1 \) so the given statement is true for n = 1

;)

correct:)

ty :) im clueless on next step :(

oh

oops and a_(k-2)
so we have also a_(k-2)<2^(k-2)

\(a_{k+1}=a_k+a_{k-1}+a_{k-2} \lt 2^k + 2^{k-1} + 2^{k-2}\)

yep good so far
do you want to try to figure out the trick into showing <2^(k+1)?

I'm trying at the moment lol, need to make the right hand side 2^{k+1}

I used what I wanted to show to help me start out
I don't know if that helps you

im getting right hand side as : \(2^{k-2}(2^2 + 2 + 1) = 2^{k-2}(7)\)

not sure how to conclude... only if that 7 was 8... :o

use 7<8

it has to be \(\ge 8 \) i think

I liked factoring out 2^(k+1) instead of 2^(k-2) but you should be able to proceed as you are

and nip is right we want <

you have your term <2^(k-2) (7)<2^(k-2) (8)=2^(k+1)

so far i have this :
\(a_{k+1}=a_k+a_{k-1}+a_{k-2} \lt 2^k + 2^{k-1} + 2^{k-2} = 2^{k-2}(7)\)

Ohhh yess !

yeah factoring out 2^(k+1) will do too.

thanks a lot that was easier than it looked initally to start wid lol xD

both ways require you to know 7<8
that is interesting

yeah. :D

hahah yes this looks easy to read quick :) thank you both :D

you're welcome (Y)