<s>[INST] Showing $1+2+\cdots+n=\frac{n(n+1)}{2}$ by induction (stuck on inductive step) This is from  this website:

Use mathematical induction to prove that 
  $$1 + 2 + 3 +\cdots+ n = \frac{n (n + 1)}{2}$$
  for all positive integers $n$.
Solution to Problem 1: Let the statement $P(n)$ be 
  $$1 + 2 + 3 + \cdots + n = \frac{n (n + 1)}{2}.$$
STEP 1: We first show that $P(1)$ is true. 
Left Side $= 1$
Right Side $= \frac{1 (1 + 1)}{2} = 1$
Both sides of the statement are equal hence $P(1)$ is true. 
STEP 2: We now assume that $P(k)$ is true
  $$1 + 2 + 3 + \cdots + k = \frac{k (k + 1)}{2}$$
  and show that $P(k + 1)$ is true by adding $k + 1$ to both sides of the above statement 
  $$
\begin{align}
1 + 2 + 3 + \cdots + k + (k + 1) &= \frac{k (k + 1)}{2} + (k + 1) \\
&= (k + 1)\left(\frac{k}{2} + 1\right) \\
&= \frac{(k + 1)(k + 2)}{2}
\end{align}
$$
  The last statement may be written as 
  $$1 + 2 + 3 + \cdots + k + (k + 1) = \frac{(k + 1)(k + 2)}{2}$$
  Which is the statement $P(k + 1)$.

My question is how in the very last line is the statement $P(k + 1)$ equal to $\frac{(k + 1)(k + 2)}{2}$. I don't get the last step.
 [/INST] We know that 
$P(k) = 1 + 2 + 3 + ... +  k$
Therefore:
$P(k+1) = 1 + 2 + 3 + ... + k + (k+1)$
By induction hypothesis we have:
$1 + 2 + 3 + ... + k + (k+1) = \frac{(k+1)(k+2)}{2}$
so
$P(k+1) = 1 + 2 + 3+...+k+(k+1) = \frac{(k+1)(k+2)}{2}$
so
$P(k+1) = \frac{(k+1)(k+2)}{2}$
By induction we now know that since this is true for one integer $k$, it is true for all integers greater than or equal to $k$.
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