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joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.0What are our tool? Are you trying to do this with induction? or do we have modular arithmetic at our disposal?

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3My favourite would be to write it as n(n+1)(n+2) + 3n(n+1) The second term is divisible by 3 and the first one is the product of 3 consecutive integers therefore also divisible by 3. The sum of 2 numbers both divisible by 3 is also divisible by 3.

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0@beginnersmind i didn't get it :/ sorry can u explain it step by step ?

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0According to induction: Put n = 1 and see are you getting what the question says..

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0@beginnersmind do not stop. Carry on.. I am just trying as @ashna is doing..

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0i know that water tell me from , to prove p(k+1) is true

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3My answer didn't use induction so I'd rather not go into a long explanation.

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0Ha ha ha ha... You knew that?? Just kidding..

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0we get k = 3M/ (K+1)(K+5) Right ?

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0Replace n by k+1 first...

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0Then Look carefully it will also be divisible by 3.. Ha ha ha ha...

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0c'mon Water i don't understand :I

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0\[= (k+1)(k+2) (k+6)\]

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0where r yu goin to substitute k = 3M/ (K+1)(K+5) ?

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0It is now 6 when I studied Induction..

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3Ok, this is how you do it with induction. First prove it for n =1 (plug it in and check if it's divisible by 3) Second assume that it's true for P(k). Using this try to prove it's also true for P(k+1) In this case I would try to prove that P(k+1)  P(k) is divisible by 3.

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0yeah .. on assuming i got k = 3M/ (K+1)(K+5) 3rd step am stuck :I

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3what does the M stand for?

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3Ah, ok, see what you did there. You said there's a number M such that P(k) = 3M

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3Ok, I'd do it slightly differently. I'd prove that the difference of P(k+1) and P(k) is divisible by 3. Then using this and the induction hypothesis it follows that P(k+1) is also divisible by 3. Does that make sense?

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3Cool :) To check, what did you get for P(k+1)  P(k) ?

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0What if we find the value of k+1 from the assumption??

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0\[k+1 = \frac{3M}{k (k+5)}\]

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0Hey I don't know how to prove that, I am just giving my Ideas which can be useless too..

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0Now this does not look like if it is divisible by 3 or not??

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0Wait, do not trust me.. Please confirm it from good source, may be I am wrong because I am not sure..

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0@beginnersmind suggest :)

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3I don't think we are going in the right direction :( Let's just start by checking if the statement holds for n = 1

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3And by let us, I mean you ;)

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus help here please..

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3Let me rephrase: What's the value of P(1)?

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0divisible by 3 , so true !

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3Ok. So we've proven that it holds for n =1. Now, what's P(k)? P(k+1)?

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3Let's forget about the induction hypothesis, just express P(k) and P(k+1) using the original definition

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0She has assumed that p(k) is divisible by 3 so 3M where M is an Integer..

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0i dont know how to do other than that :I @beginnersmind

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3I mean, the same way you plugged in n = 1 to get P(1) you can plug in k to get P(k)

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0Just replace n by k there..

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0okay so k(k+1)(k=5) = divisible by 3 yu meant this ?

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0p(k) = .......................

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3I mean P(k) = k(k+1)(k+5) We'll use the fact that it's divisible by 3 later, but not yet.

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0okay , now how to proceed ?

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3So the same way, what is P(k+1)?

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3We want to look at P(k+1)  P(k)

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3This is the difference between two consecutive terms. For example the difference between the second term and the first term is P(2)P(1)

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3If P(2)  P(1) is divisible by 3 and P(1) is divisible by 3 then so is P(2). Do you see why?

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3Let's try to write out P(k+1)  P(k) and see if it's divisible by 3

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3You already calculated P(k+1) and P(k) separately so just take the difference and see if you can write it in some nice way.

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0It p(1) = 6 p(2) = 8 Then : p(2)  p(1) = 2

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0okay i've got (k+1)[(3k+12)]

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.0This is divisible by 3 now...

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3That's nice. I'd rewrite it as 3(k+1)(k+4) just to emphasize that it's divisible by 3.

ashna
 2 years ago
Best ResponseYou've already chosen the best response.0right YAY :D ty @beginnersmind and @waterineyes

beginnersmind
 2 years ago
Best ResponseYou've already chosen the best response.3Not finished yet. But almost :)
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