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

Question

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 $\forall n\in \mathbb{N}$.

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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?

Answer 1

Nice work! If you want to take a bit more time, you can note that $$\frac{k^2+3k+2}2=\frac{k^2+2k+k+2}2=\frac{k(k+2)+1(k+2)}2=\frac{(k+1)(k+2)}2.$$

Answer 2

Q1: No problems, that's the way induction works.

Q2: go back one step:

$$k(k+1)+2k+2=k(k+1)+2(k+1)=(k+1)(k+2)$$

Answer 3

No, all the steps look pretty good; as for the unknown $[\text{step}]$, that is just factoring - although, it is easier to notice that $$ \frac{k(k+1) + 2k + 2}{2} = \frac{k(k+1) + 2 (k+1)}{2} = \frac{(k+1)(k+2)}{2} . $$