How does this prove that P(k) of every k is true? This is an example from my textbook. I'm very rusty with simplifying algebraic expression so i hope you'll forgive me for that.
The textbook says there are two rules to Mathematical Induction:
1) We must first prove that $P(1)$ is true.
2) We must then assume that $P(k)$ is true and prove that $P(k+1)$ is true.
Show that if n is a positive integer, then
$1 + 2+· · ·+n =\frac{ n(n + 1)}
2$
For the inductive hypothesis we assume that P(k) holds for an arbitrary
positive integer k. That is, we assume that
$1 + 2+· · ·+k = \frac{k(k + 1)}  2$
.
Under this assumption, it must be shown that P(k + 1) is true, namely, that
$1 + 2+· · ·+k + (k + 1) =\frac {(k + 1)((k + 1) + 1)}
2= \frac{(k + 1)(k + 2)}
2$ 
is also true. When we add $k + 1$ to both sides of the equation in P(k), we obtain
$1 + 2+· · ·+k + (k + 1)
 =\frac{k(k + 1)}
2
+ (k + 1)
= \frac{k(k + 1) + 2(k + 1)}
2$
$= \frac{(k + 1)(k + 2)}
2$
.
This last equation shows that P(k + 1) is true under the assumption that P(k) is true. This
completes the inductive step.
My question is how this proves that $P(k+1)$ is true? Also, why does textbook add $k+1$ to both sides of the equation? 

You are corect that what we want to show is that assuming:  $$1 + 2 + \cdots + k = \frac{k(k+1)}{2} \hspace{40mm} (1)$$ 
it is also true that: $$1 + 2 + \cdots + k + (k+1) = \frac{(k+1)(k+2)}{2}. \hspace{10mm} (2)$$
So to do this we are aloud to manipulate the expression we know to be true. Now since we know $(1)$ we area allowed to manipulate it. The idea is it's easy to transform the left side of $(1)$ into the left side of $(2)$ by adding $k + 1$. This is why the textbook adds $k +1$ to both sides of equation $(1)$.
Now when we do this we get $$1 + 2 + \cdots + k + (k+1) = \frac{k(k+1)}{2} + (k+1) = \frac{k(k+1)}{2} + \frac{2(k+1)}{2} = \frac{(k+2)(k+1)}{2}.$$
So now we have shown that, assuming (1) is true (2) must also be true, which is why this proves that P(k+1) is true. 
EDIT: To adress the comment, it is from the fact that $\frac{2}{2} = 1$: 
$$\frac{k(k+1)}{2} + (k+1) = \frac{k(k+1)}{2} + (k+1)\cdot1 = \frac{k(k+1)}{2} + (k+1)\cdot\frac{2}{2}= \frac{k(k+1)}{2} + \frac{2(k+1)}{2}$$
