Why is it that when we derive, say, a closed form to the series $1+2+\dotsc+n$, which is $\frac{n(n+1)}{2}$, it is almost habitual to do a proof by induction on the equation? If the $n$ was arbitrarily chosen at the beginning of the series, and then the closed form was deduced, why bother with PMI to see if it works for all $n \in \mathbb{N}$? The $n$ was arbitrary, so it should work for all $n$ by default.
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What if I told you that $1+2+3+\ldots+n$ was equal to $n^2$? The $n$ here is arbitrary, so it should work for all $n$. If you're really a skeptic, you can check it for $n=1$ and even $n=0$. Trust me, I deduced this the other night - it's definitely correct.
The point here is that we do not initially know whether two things are equal. If you wish to show that $1+2+3+\ldots+n$ equals $\frac{n(n+1)}2$ for all $n$, you need to come up with some proof of this and induction is one easy method to do this. Of course, other means to prove things like this exist, and may line up better with how you would go about deriving the result if you didn't know it, but some verification is necessary, and induction is a nice straightforwards method to use.
Milo Brandt
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You missed the part(at the beginning) of my post where I said we already derived the closed form for the finite series, so my question is what is the point of pmi if we already derived the closed form and the $n$ was arbitrarily chosen prior to the derivation. – J. Dunivin Nov 04 '15 at 02:11
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@BenedictVoltaire My last sentence alludes to this. What precisely do you mean by "derivation"? Because it might just be another proof (or possibly induction in disguise) – Milo Brandt Nov 04 '15 at 02:13
$\mathrm{log}_2(3)\cdot \mathrm{log}_3(4)\cdot \mathrm{log}(5) \cdot \dotsc \cdot \mathrm{log}_n(n+1)$
and derive the closed form $ \mathrm{log}_2 (n+1)$ that equals the above product, would it be necessary to prove by induction that the equation holds for all $n$?
@JohnDouma No.
– J. Dunivin Nov 04 '15 at 02:01