Applying the principle of mathematical Induction, prove that

Question

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?

Answer 1

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 $m \in \mathbb{N}$

Answer 2

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.

Answer 3

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}