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Prove that P(n) : n(n+1)(n+5) is divisible by 3

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What are our tool? Are you trying to do this with induction? or do we have modular arithmetic at our disposal?

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Other answers:

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

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