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Applying the principle of mathematical Induction, prove that $n(n+1)(2n+1)$ is divisible by $6$ for all $n\in N$

My Attempt Let $$P(n)=n(n+1)(2n+1)$$ When $n=1$, then $$P(1)=6$$ Which is divisible by $6$. Hence $P(1)$ is true. Suppose $P(k)$ is true for some $K\in N$ $$P(k)=k(k+1)(2k+1)$$ We have to prove that $P(k+1)$ is true whenever $P(k)$ is true. Thus, $$P(k+1)=(k+1)(k+2)(2k+3)$$ $$=(k+1)(k+2)(2k+1+2)$$

How do I complete the proof?

pi-π
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3 Answers3

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Set up is everything with these proofs!

In particular, you need to clearly define your terms, and you don't do that:

When you write:

$$P(n) = n(n+1)(2n+1)$$

you define $P(n)$ as a number

But when you say

Suppose $P(k)$ is true for some $k$ ...

you treat $P(k)$ as a claim.

I recommend that you define $P(n)$ as the claim that the number $n(n+1)(2n+1)$ is divisible by $6$.

So, when you say that 'Suppose $P(k)$ is true for some $k$', you thereby assume the claim that $k(k+1)(2k+1)$ is divisible by $6$, i.e. that $k(k+1)(2k+1)=6m$ for some $m \in \mathbb{N}$

Bram28
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  • the first one is base step and $P(k)$ is for inductive hypothesis. – pi-π May 21 '18 at 17:42
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    @blue_eyed_... But what is $P(k)$? Is it a number or a claim? – Bram28 May 21 '18 at 17:43
  • that's a claim. – pi-π May 21 '18 at 17:44
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    @blue_eyed_... then what is meant by "$P(k)\color{red}{=}k(k+1)(2k+1)$"? If $P(k)$ is a claim and not a number, you have a claim on the left hand of the "equals sign" and a number on the right hand of the "equals" sign – JMoravitz May 21 '18 at 17:45
  • @blue_eyed_... OK, good. So, you should start with $k(k+1)(2k+1)=6m$, rather than with $P(k)=k(k+1)(2k+1)$ – Bram28 May 21 '18 at 17:46
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You may complete the proof by a simple factorisation of $P(k+1)-P(k)$ to prove the difference is divisible by $6$.

Indeed, we have \begin{align} P(k+1)-P(k)&=(k+1)(k+2)(2k+3)-k(k+1)(2k+1) \\ &=(k+1)\bigl((k+2)(2k+1+2)-k(2k+1)\bigr)\\ &=(k+1)\bigl((k+2-k)(2k+1)+2(k+2)\bigr) \\ &=(k+1)\bigl(2(2k+1)+2(k+2)\bigr) =(k+1)(6k+6)=\color{red}{6(k+1)^2}. \end{align}

Bernard
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Your base case looks good. Now, assume true for $n=k$.

$$k(k+1)(2k+1)=6m\to2k^3+3k^2+k=6m$$

Then show for $n=k+1$,

$$(k+1)(k+2)(2k+3)=2k^3+9k^2+13k^2+6$$

Thus $$P(k+1)-P(k)=6k^2+12k+6=6(k+1)^2$$

Since we assumed $P(k)=6m$, we now see that: $$P(k+1)=6(m+(k+1)^2)$$ and therefore true. Then write your conclusion step.

Rhys Hughes
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