anonymous
  • anonymous
Prove that P(n) : n(n+1)(n+5) is divisible by 3
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@waterineyes
anonymous
  • anonymous
What are our tool? Are you trying to do this with induction? or do we have modular arithmetic at our disposal?
anonymous
  • anonymous
induction

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More answers

beginnersmind
  • beginnersmind
My 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.
anonymous
  • anonymous
@beginnersmind i didn't get it :/ sorry can u explain it step by step ?
anonymous
  • anonymous
According to induction: Put n = 1 and see are you getting what the question says..
anonymous
  • anonymous
@beginnersmind do not stop. Carry on.. I am just trying as @ashna is doing..
anonymous
  • anonymous
i know that water tell me from , to prove p(k+1) is true
beginnersmind
  • beginnersmind
My answer didn't use induction so I'd rather not go into a long explanation.
anonymous
  • anonymous
Ha ha ha ha... You knew that?? Just kidding..
anonymous
  • anonymous
we get k = 3M/ (K+1)(K+5) Right ?
anonymous
  • anonymous
Replace n by k+1 first...
anonymous
  • anonymous
yeah did , then ?
anonymous
  • anonymous
Then Look carefully it will also be divisible by 3.. Ha ha ha ha...
anonymous
  • anonymous
c'mon Water i don't understand :I
anonymous
  • anonymous
\[= (k+1)(k+2) (k+6)\]
anonymous
  • anonymous
okay
anonymous
  • anonymous
Really??
anonymous
  • anonymous
where r yu goin to substitute k = 3M/ (K+1)(K+5) ?
anonymous
  • anonymous
Wait...
anonymous
  • anonymous
okay
anonymous
  • anonymous
It is now 6 when I studied Induction..
anonymous
  • anonymous
*6 years..
beginnersmind
  • beginnersmind
Ok, 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.
anonymous
  • anonymous
yeah .. on assuming i got k = 3M/ (K+1)(K+5) 3rd step am stuck :I
beginnersmind
  • beginnersmind
what does the M stand for?
beginnersmind
  • beginnersmind
Ah, ok, see what you did there. You said there's a number M such that P(k) = 3M
anonymous
  • anonymous
M = divisible by 3
beginnersmind
  • beginnersmind
Ok, 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?
anonymous
  • anonymous
yes :)
beginnersmind
  • beginnersmind
Cool :) To check, what did you get for P(k+1) - P(k) ?
anonymous
  • anonymous
What if we find the value of k+1 from the assumption??
anonymous
  • anonymous
\[k+1 = \frac{3M}{k (k+5)}\]
anonymous
  • anonymous
|dw:1355641516686:dw|
anonymous
  • anonymous
Hey I don't know how to prove that, I am just giving my Ideas which can be useless too..
anonymous
  • anonymous
Now this does not look like if it is divisible by 3 or not??
anonymous
  • anonymous
yeah .. :P
anonymous
  • anonymous
Wait, do not trust me.. Please confirm it from good source, may be I am wrong because I am not sure..
anonymous
  • anonymous
@beginnersmind suggest :)
beginnersmind
  • beginnersmind
I don't think we are going in the right direction :( Let's just start by checking if the statement holds for n = 1
anonymous
  • anonymous
okay !
beginnersmind
  • beginnersmind
And by let us, I mean you ;)
anonymous
  • anonymous
@UnkleRhaukus help here please..
beginnersmind
  • beginnersmind
Let me rephrase: What's the value of P(1)?
anonymous
  • anonymous
12
anonymous
  • anonymous
divisible by 3 , so true !
beginnersmind
  • beginnersmind
Ok. So we've proven that it holds for n =1. Now, what's P(k)? P(k+1)?
anonymous
  • anonymous
p(k) = 3M / (k+1)(k+5)
beginnersmind
  • beginnersmind
Let's forget about the induction hypothesis, just express P(k) and P(k+1) using the original definition
anonymous
  • anonymous
She has assumed that p(k) is divisible by 3 so 3M where M is an Integer..
anonymous
  • anonymous
i dont know how to do other than that :I @beginnersmind
beginnersmind
  • beginnersmind
I mean, the same way you plugged in n = 1 to get P(1) you can plug in k to get P(k)
anonymous
  • anonymous
Just replace n by k there..
anonymous
  • anonymous
okay so k(k+1)(k=5) = divisible by 3 yu meant this ?
anonymous
  • anonymous
p(k) = .......................
beginnersmind
  • beginnersmind
I mean P(k) = k(k+1)(k+5) We'll use the fact that it's divisible by 3 later, but not yet.
anonymous
  • anonymous
okay , now how to proceed ?
beginnersmind
  • beginnersmind
So the same way, what is P(k+1)?
anonymous
  • anonymous
P(k+1) = (k+1)(k+2)(k+6)
beginnersmind
  • beginnersmind
right
anonymous
  • anonymous
and now ?
beginnersmind
  • beginnersmind
We want to look at P(k+1) - P(k)
anonymous
  • anonymous
why ?
beginnersmind
  • beginnersmind
This is the difference between two consecutive terms. For example the difference between the second term and the first term is P(2)-P(1)
anonymous
  • anonymous
okay
beginnersmind
  • beginnersmind
If P(2) - P(1) is divisible by 3 and P(1) is divisible by 3 then so is P(2). Do you see why?
anonymous
  • anonymous
yes i do
beginnersmind
  • beginnersmind
Let's try to write out P(k+1) - P(k) and see if it's divisible by 3
beginnersmind
  • beginnersmind
You 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.
anonymous
  • anonymous
It p(1) = 6 p(2) = 8 Then : p(2) - p(1) = 2
anonymous
  • anonymous
okay i've got (k+1)[(3k+12)]
anonymous
  • anonymous
This is divisible by 3 now...
beginnersmind
  • beginnersmind
That's nice. I'd rewrite it as 3(k+1)(k+4) just to emphasize that it's divisible by 3.
anonymous
  • anonymous
right YAY :D ty @beginnersmind and @waterineyes
beginnersmind
  • beginnersmind
Not finished yet. But almost :)
anonymous
  • anonymous
i can do the rest ty :)
beginnersmind
  • beginnersmind
Ok :)

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