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I saw some video where it needs to prove $1+2+\dots+n=\frac{n(n+1)}{2}$ inductively. So it has to be true if $k=1$ and $k+1$ are true.

So, for $k=1$:

$$1=\frac{1(1+1)}{2}=\frac{1(2)}{2}=\frac{2}{2}=1$$

it is valid.

For $k+1$ here is the proof he does:

$$ \begin{align} 1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2} &(1)\\ \frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2} &(2)\\ &=\frac{k^2+2k+k+2}{2}&(3)\\ &=\frac{k^2+3k+2}{2}&(4)\\ &=\frac{(k+1)(k+2)}{2}&\text{factoring (4)} \end{align}$$

Therefore this formula is valid for $k+1$.


But is this true? I think not. He is just undoing what he have just done. To prove it I think I need to do this:

$$ \begin{align} 1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2}\\ \frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2}\\ \frac{k(k+1)+2(k+1)}{2}&=\frac{(k+1)(k+2)}{2}\\ \frac{(k+1)(k+2)}{2}&=\frac{(k+1)(k+2)}{2}\\ (k+1)(k+2)&=(k+1)(k+2)\\ k+1&=k+1\\ k&=k\\ 0&=0\text{ or }1=1 \end{align}$$

Therefore, $k+1$ is valid in this formula.

Is this right or am I making a mistake somewhere?

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  • $\begingroup$ Why $k = k$ implies that $0=0$ or $1=1$? That's is true for every $k$. $\endgroup$ – Luísa Borsato Jan 27 '17 at 11:45
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    $\begingroup$ It's odd to begin with an equality like this. The correct way would be $1+...+k+k+1 = (1+...+k)+k+1 = (k(k+1)/2) + k+1$ etc. $\endgroup$ – Max Jan 27 '17 at 11:47
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The second way is correct but I would be more carefull and use like this:

$$1+2+...+k+(k+1)=\frac{k(k+1)}{2}+k+1=\frac{k^2+3k+2}{2}=\frac{(k+1)(k+2)}{2}$$

P.s: Not use the equality in the first place. The problem using the equality is that you need guarantee equivalence in every step what is not so easy in many problems.

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Proving for sum =k+1

RhS= $\dfrac {(k)(k+1)}{2} + k + 1 $

Which gives

=$\dfrac {(k+1)(k+2)}{2}$

= $\dfrac {([k+1])([k+1] +1)}{2}$

Which means it hold for n and also n+1 !

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This proof is at least unlucky written down.

Normally, we start with the claim for $n$ (Here $1+2+\cdots n=\frac{n(n+1)}{2}$) and then proof the claim for $n+1$.

The correct way is to start with $1+2+\cdots n+n+1=\frac{n(n+1)}{2}+n+1$ (we assume that the formula is correct for $n$) and show that this is equal to $\frac{(n+1)(n+2)}{2}$ by simply adding the two fractions.

It is unusual to start with the aim of the induction step and then going back to show that it is valid.

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You don't to begin using equality!! You must to arrive in this... $1+2+⋯+k+k+1= $(here you use the induction hipoteses) Than...We start correct:

$$1+2+⋯+k+k+1=**induction hipoteses(ih)**\frac{k(k+1)}{2}**finish(ih)**+**hereyourepeatlikeabove**k+1$$

I hope I've helped.

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Sua ultima afirmaçao esta errada. Parece que voce nao entendeu bem a "hipotese de induçao". "Se vale para k+1 entao valera para k".

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    $\begingroup$ People use english here! Not portuguese! $\endgroup$ – Arnaldo Jan 27 '17 at 11:50
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    $\begingroup$ Why? ok. I can translate for you. $\endgroup$ – Cappellesso Jan 27 '17 at 11:52
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    $\begingroup$ You don't need translate for me because I speak portuguese, but we have people around the world here and most of them don't understand portuguese. $\endgroup$ – Arnaldo Jan 27 '17 at 11:55
  • $\begingroup$ I understand, I tried explain it better. Sorry for my English. $\endgroup$ – Cappellesso Jan 27 '17 at 12:16

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