A proposed proof by induction of $1+2+\ldots+n=\frac{n(n+1)}{2}$ 
Prove: $\displaystyle 1+2+\ldots+n=\frac{n(n+1)}{2}$.

Proof
When $n=1,1=\displaystyle \frac{1(1+1)}{2}$,equality holds.
Suppose when $n=k$, we have $1+2+\dots+k=\frac{k(k+1)}{2}$
When $n = k + 1$:
\begin{align}
1+2+\ldots+k+(k+1) &=\frac{k(k+1)}{2}+k+1 =\frac{k(k+1)+2k+2}{2}\\
&=\frac{k^2+3k+2}{2}\\
\text{[step]}&=\displaystyle\frac{(k+1)(k+2)}{2}=\displaystyle\frac{(k+1)((k+1)+1)}{2}
\end{align}
equality holds.
So by induction, the original equality holds $\forall n\in \mathbb{N}$.

Question 1: any problems in writing?
Question 2: Why [step] happen to equal? i.e., why does $k^2+3k+2=(k+1)(k+2)$ hold?

Q1: No problems, that's the way induction works.
Q2: go back one step:
$$k(k+1)+2k+2=k(k+1)+2(k+1)=(k+1)(k+2)$$
