Question

prove that n(n+1)(2n+1) is a multiple of 6

Answer 1

in recomend expanding the brackets first

Answer 2

expand the first bracket and what do u get?

Answer 3

This is more like a proof by induction, where you have to plug in numbers for n and see if hey equate to 6.

Answer 4

So you have \(\sf n(n+1)(2n+1)=6\), which value of \(\sf n\) could you pplug into the function on the left to make this true?

Answer 5

We need to prove this by induction

\(2n^3+3n^2+n\)

This is for sure divisible by \(6\) for \(n=1\).

So we assume it is divisible for some \(n\) then prove its true for \(n+1\)

Plug in \(n+1\) to the above equation and we get\(2n^3+6n^2+3n^2+12n+n+6=[2n^3+3n^2+n]+6(n^2+2n+1)
\)

The stuff in the \([]\) is divisible by \6\) by assumption and obviously the stuff after that is divisible by \(6\).

Thus by induction we have proved the desired result.

Answer 6

The stuff in the \([]\) is divisible by \(6\)*

Answer 7

@adihsai does this make sense?

Answer 8

You could also let \(\sf n = 1\ , \ 2 \ , \ 3 \ , ...\) Right? Or is it necessary to expand? O_o

Answer 9

well I am sure there is some other way you could construct it, but if you are trying to show its true for all n then induction is usually the easiest

Answer 10

n(n+1) is trivially divisible by 2.
Prove that n(n+1)(2n+1) is divisible by 3 and you're done.

Answer 11

Alternatively,\[\frac{n(n + 1) (2n + 1)}{6} = \sum n^2 \in \mathbb Z\]

Answer 12

You can also prove that n(n+1)(2n+1) is even for all n

if n is even, then n(n+1)(2n+1) is even
if n is odd, then n+1 is even making n(n+1)(2n+1) even

so n(n+1)(2n+1) is divisible by 2

Answer 13

That way makes more sense to me, in my opinion. @Owlfred

Answer 14

You can use modular arithmetic to prove n(n+1)(2n+1) is divisible by 3


if n = 0 (mod 3), then

n = 0 (mod 3)
n+1 = 1 (mod 3)
2n+1 = 1 (mod 3)


since n = 0 (mod 3), this means n is divisible by 3. So n(n+1)(2n+1) is divisible by 3


-------------------

if n = 1 (mod 3), then

n = 1 (mod 3)
n+1 = 2 (mod 3)
2n+1 = 2*1+1 = 2+1 = 3 = 0 (mod 3)

since 2n+1 = 0 (mod 3), this means 2n+1 is divisible by 3. So n(n+1)(2n+1) is divisible by 3


-------------------

if n = 2 (mod 3), then

n = 2 (mod 3)
n+1 = 3 = 0 (mod 3)
2n+1 = 2*2+1 = 5 = 2 (mod 3)

since n+1 = 0 (mod 3), this means n+1 is divisible by 3. So n(n+1)(2n+1) is divisible by 3

------------------

Any other integral values of n will either land on n = 0 (mod 3), n = 1 (mod 3) or n = 2 (mod 3). So this proves n(n+1)(2n+1) is divisible by 3

Answer 15

I am almost willing to be my left pinky that the user is working on induction in class:)

Answer 16

\[n(n+1)(2n+1) = n(n+1)(2n+4-3) = n(n+1)(n+2) - 3n(n+1)\]

Answer 17

If they are working on mathematical induction, I sincerely hope whoever is grading the assignment is open to other ideas.

Answer 18

they're probably strict, but some teachers welcome tackling problems from multiple angles

Answer 19

Now, if the REAL problem statement is "prove by mathematical induction that n(n+1)(2n+1) is a multiple of 6", then there is a good argument for strictness.

Answer 20

but it is probably just an induction problem (well the teacher intended it to be)

Answer 21

Just seems like after I have proved it and am trying to get feedback from the user by asking if they understand, and then everything gets cluttered with segments of proofs, and comments...Seems unproductive and might lead to confusion. This is just my opinion. Good luck everyone. pz.

Answer 22

The absent of fact that it should be proved using mathematical induction doesn't mean it is not good idea to use mathematical induction.

Given statement simply need to be proved.

Answer 23

If OP's teacher doesn't accept proof by mathematical induction, it's OP fault for not being specified on his question here.

Answer 24

@Owlfred how could we prove that?