Question on proof of $1+2+\dots+n=\frac{n(n+1)}{2}$ by induction. 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:

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

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:

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

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

Is this right or am I making a mistake somewhere?
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.
Sua ultima afirmaçao esta errada. Parece que voce nao entendeu bem a "hipotese de induçao". "Se vale para k+1 entao valera para k".