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

- anonymous

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

- chestercat

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- anonymous

- anonymous

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

- anonymous

induction

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## More answers

- 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

@beginnersmind i didn't get it :/ sorry
can u explain it step by step ?

- anonymous

According to induction:
Put n = 1 and see are you getting what the question says..

- anonymous

@beginnersmind do not stop.
Carry on..
I am just trying as @ashna is doing..

- anonymous

i know that water tell me from ,
to prove p(k+1) is true

- beginnersmind

My answer didn't use induction so I'd rather not go into a long explanation.

- anonymous

Ha ha ha ha...
You knew that??
Just kidding..

- anonymous

we get k = 3M/ (K+1)(K+5) Right ?

- anonymous

Replace n by k+1 first...

- anonymous

yeah did , then ?

- anonymous

Then Look carefully it will also be divisible by 3..
Ha ha ha ha...

- anonymous

c'mon Water i don't understand :I

- anonymous

\[= (k+1)(k+2) (k+6)\]

- anonymous

okay

- anonymous

Really??

- anonymous

where r yu goin to substitute k = 3M/ (K+1)(K+5) ?

- anonymous

Wait...

- anonymous

okay

- anonymous

It is now 6 when I studied Induction..

- anonymous

*6 years..

- 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

yeah .. on assuming i got k = 3M/ (K+1)(K+5)
3rd step am stuck :I

- beginnersmind

what does the M stand for?

- beginnersmind

Ah, ok, see what you did there. You said there's a number M such that P(k) = 3M

- anonymous

M = divisible by 3

- 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

yes :)

- beginnersmind

Cool :)
To check, what did you get for P(k+1) - P(k) ?

- anonymous

What if we find the value of k+1 from the assumption??

- anonymous

\[k+1 = \frac{3M}{k (k+5)}\]

- anonymous

|dw:1355641516686:dw|

- anonymous

Hey I don't know how to prove that, I am just giving my Ideas which can be useless too..

- anonymous

Now this does not look like if it is divisible by 3 or not??

- anonymous

yeah .. :P

- anonymous

Wait, do not trust me..
Please confirm it from good source, may be I am wrong because I am not sure..

- anonymous

@beginnersmind suggest :)

- 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

okay !

- beginnersmind

And by let us, I mean you ;)

- anonymous

@UnkleRhaukus help here please..

- beginnersmind

Let me rephrase:
What's the value of P(1)?

- anonymous

12

- anonymous

divisible by 3 , so true !

- beginnersmind

Ok. So we've proven that it holds for n =1.
Now, what's
P(k)?
P(k+1)?

- anonymous

p(k) = 3M / (k+1)(k+5)

- beginnersmind

Let's forget about the induction hypothesis, just express P(k) and P(k+1) using the original definition

- anonymous

She has assumed that p(k) is divisible by 3 so 3M where M is an Integer..

- anonymous

i dont know how to do other than that :I
@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

Just replace n by k there..

- anonymous

okay so k(k+1)(k=5) = divisible by 3
yu meant this ?

- anonymous

p(k) = .......................

- 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

okay , now how to proceed ?

- beginnersmind

So the same way, what is P(k+1)?

- anonymous

P(k+1) = (k+1)(k+2)(k+6)

- beginnersmind

right

- anonymous

and now ?

- beginnersmind

We want to look at P(k+1) - P(k)

- anonymous

why ?

- 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

okay

- 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

yes i do

- beginnersmind

Let's try to write out P(k+1) - P(k) and see if it's divisible by 3

- 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

It p(1) = 6
p(2) = 8
Then : p(2) - p(1) = 2

- anonymous

okay i've got (k+1)[(3k+12)]

- anonymous

This is divisible by 3 now...

- beginnersmind

That's nice. I'd rewrite it as 3(k+1)(k+4) just to emphasize that it's divisible by 3.

- anonymous

right YAY :D
ty @beginnersmind and @waterineyes

- beginnersmind

Not finished yet.
But almost :)

- anonymous

i can do the rest ty :)

- beginnersmind

Ok :)

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