Induction Question (1 Viewer)

rand_althor · Jun 5, 2015

Hey could someone please help with the following induction question?

Thanks!

VBN2470 · Jun 5, 2015

Step (I): Base Case $$S_1=0+1+8=9$$ which is divisible by 3.

Step (II): Hypothesis (or assumption) $$S_n=3m$$ for some $$m\in\mathbb{Z}$.$

Step (III): Inductive Step: Prove that $S_{n+1}$ is divisible by 3.

$$S_{n+1}\\=n^3+(n+1)^3+(n+2)^3\\=S_{n}-(n-1)^3+(n+2)^3\\=3m-n^3+3n^2-3n+1+n^3+12n^2+6n+8=3m+3(5n^2+n+3)=3p$, $p\in\mathbb{Z}$$

Hence, $S_n$ is divisible by 3 for all $$n\in\mathbb{N}$$ .

InteGrand · Jun 5, 2015

And if we don't feel like expanding, we should be allowed to straightaway say in the Induction Step that $3|((n+2)^3 - (n-1)^3)$ , due to the factorisation of $(n+2)^3 - (n-1)^3$ as $(n+2-(n-1))\times (\text{etc.}) = 3\times (\text{etc.})$ , where $\text{etc.}$ is an integer.

rand_althor · Jun 5, 2015

VBN2470 said:
Step (I): Base Case $$S_1=0+1+8=9$$ which is divisible by 3.

Step (II): Hypothesis (or assumption) $$S_n=3m$$ for some $$m\in\mathbb{Z}$.$

Step (III): Inductive Step: Prove that $S_{n+1}$ is divisible by 3.

$$S_{n+1}\\=n^3+(n+1)^3+(n+2)^3\\=S_{n}-(n-1)^3+(n+2)^3\\=3m-n^3+3n^2-3n+1+n^3+12n^2+6n+8=3m+3(5n^2+n+3)=3p$, $p\in\mathbb{Z}$$

Hence, $S_n$ is divisible by 3 for all $$n\in\mathbb{N}$$ .

Thank you!

InteGrand said:
And if we don't feel like expanding, we should be allowed to straightaway say in the Induction Step that $3|((n+2)^3 - (n-1)^3)$ , due to the factorisation of $(n+2)^3 - (n-1)^3$ as $(n+2-(n-1))\times (\text{etc.}) = 3\times (\text{etc.})$ , where $\text{etc.}$ is an integer.

I tried to factorise that, but I ended up with $3k^2+9k+7$ . Did I make a mistake? If not, how would you proceed from there?

InteGrand · Jun 5, 2015

rand_althor said:
Thank you!

I tried to factorise that, but I ended up with $3k^2+9k+7$ . Did I make a mistake? If not, how would you proceed from there?

$The factorisation of $(k+2)^3 - (k-1)^3$ is only important insofar as it allows us to quickly see that it is divisible by 3 (because $a^3 - b^3=(a-b)(a^2 + ab + b^2)$, but it's really just the $(a-b)$ factor that's important for us here). But if you're interested, the full factorisation was $(k+2)^3 - (k-1)^3 = ((k+2)-(k-1))\times ((k+2)^2 + (k+2)(k-1) + (k-1)^2)=3\times (3k^2 + 3k+3)$.$

rand_althor · Jun 5, 2015

InteGrand said:
$The factorisation of $(k+2)^3 - (k-1)^3$ is only important insofar as it allows us to quickly see that it is divisible by 3 (because $a^3 - b^3=(a-b)(a^2 + ab + b^2)$, but it's really just the $(a-b)$ factor that's important for us here). But if you're interested, the full factorisation was $(k+2)^3 - (k-1)^3 = ((k+2)-(k-1))\times ((k+2)^2 + (k+2)(k-1) + (k-1)^2)=3\times (3k^2 + 3k+3)$.$

Ah I see my mistake now. Thanks for the help!

Induction Question (1 Viewer)

rand_althor

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VBN2470

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InteGrand

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rand_althor

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InteGrand

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rand_althor

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