Proof by $PMI$ of a formula for the sum of the first $n$ terms of a given sequence I am working exercices on the Principle of Mathematical Induction and I am not sure how to get out of some dead end I reach.
The terms of a sequence $\lbrace a_1, a_2,\ldots, a_{n-1}, a_n\rbrace$ are described as $a_{n+1} = a_n+r$ such that $a_1=a$ and $a,r\in\Bbb R^+$.
I want to show by $PMI$ that the sum $S_n$ of the first $n$ terms of the sequence is given by
$$S_n = an + \frac {n(n-1)}2 r$$
Base Case $P(1)$
Let $n=1$, then
$$S_1 = a_1 = a(1)+\frac 02 r = a$$
is true and I often do $n=2$ for good mesure, so
$$S_2 = a(2)+ \frac 22r = a_1 + a_2 = a+(a+r) = 2a+r$$
is also true so it the inductive step can be taken.
Induction Hypothesis $P(k)$
For $n=k$ assume that $S_k = ak + \frac {k(k-1)}2 r$ is true
Inductive Step $P(k)\Rightarrow P(k+1)$
I think my goal is to show that $S_k+(a+r)=S_{k+1}$ so I write
$$S_k+(a+r)=S_{k+1}$$
$$ak+\frac{k(k-1)}2 r + (a+r) = a(k+1)+\frac {(k+1)k}2 r$$
$$a(k+1) + \left(\frac{k(k-1)}2 +1\right)r = a(k+1) + \left(\frac{k(k+1)}2\right)r$$
and this is where it dies on me since
$$\left(\frac{k(k-1)}2 +1\right)\ne \left(\frac{k(k+1)}2\right)$$
I wonder if the problem comes from either my understanding of the $PMI$ lacking, my interpretation of the given sequence or if it is just me doing the algebra too fast and overlooking some detail even while going over the problem again?
As always, I really appreciate that you take some of your time to answer or comment, thank you very much, it helps a lot!
 A: To prove everything in one shot, try arguing as follows:
$S_{n+1} = \sum_{i=1}^{n+1}a_i = (a_{n+1} + \cdots + a_2) + a_1 = ((a_n+r) + \cdots + (a_1+r))+ a = (a_n+\cdots +a_1) + nr + a = S_n + nr + a$.
Now you can apply the inductive hypothesis $S_n = an + \frac{n(n-1)}{2}r$, after which you arrive at the desired expression for $S_{n+1}$ with little extra effort.
The reason you ran into an algebraic problem is you attempted to show $S_k + (a+r) = S_{k+1}$ when instead what you needed was that $S_k + (a+kr) = S_{k+1}$.
A: Even though this problem is solved much quicker by following joeb's argument (see answers), I wanted to include to this post the correct solution I found by following my initial idea, while leaving the original post as it is for others that might find this useful in the future.
So here it goes, 
Proposition $(\forall n \in \Bbb Z^+)P(n)$
The sum $S_n$ of the first $n$ terms of the sequence ${a_n}$ is given by
$$S_n = an + \frac {n(n-1)}2 r$$
Base Case $P(1)$
Let $n=1$, then we see
$$S_1 = a_1 = a(1)+\frac 02 r = a$$ 
is true hence, the inductive step can be taken.
Induction Hypothesis $P(k)$
For $n=k$, we assume 
$$S_k=ka+\frac{k(k-1)}2r$$
is true.
Inductive step $P(k)\rightarrow P(k+1)$
If we first take a look at $S_{k+1}$ and compare it to $S_k$ 
$$ \begin{align}
S_k & = ka+\frac{k(k-1)}2r = ak + \frac{k^2r}2-\frac{kr}2 \\
S_{k+1} & =(k+1)a+\frac{k(k+1)}2r = ak+a + \frac{k^2r}2+\frac{kr}2 \\
\end{align}$$
we can observe that both differ by $(a+kr)$ hence, showing $S_{k+1}=S_k +(a+kr)$ is the goal.
$$\begin{align}
S_{k+1} & = S_k +(a+kr) \\
(k+1)a+\frac{k(k+1)}2r & = ka+\frac{k(k-1)}2r + (a+kr) \\
& = ak+a + \frac{k^2r}2-\frac{kr}2 +kr \\
& = (k+1)a+\frac{k(k+1)}2r \\
\end{align}$$
$$\tag*{$\blacksquare$}$$
Conclusion
$$P(1)\rightarrow P(k) \rightarrow P(k+1) \rightarrow \left(\forall n \in \Bbb Z^+\right)P(n)$$
