Proof by induction that sum of terms $a_i$ with $a_i = b + i \cdot c$ is $(n+1) \frac{a_0 + a_n}{2}$ So I have this problem: 

Let the finite sequence $a_0,a_1,\ldots,a_n$ be defined by $a_i = b + i\cdot c$. Prove by induction on $n$ that the sum of the terms in the sequence is $\displaystyle(n+1)\frac{a_0 + a_n}{2}$. (Hint: in the base case, $n = 0$ and so $a_0$ is equal to $a_n$. For the induction case, note that the sum for $n+1$ is equal to the sum for $n$ plus the one new term $a_{n+1}$)

So for the base case I have that $a_0$ is equal to $b$ and so $a_n$ is also equal to $b$. When I plug that into the equation $\displaystyle(n+1)\frac{a_0 + a_n}{2}$ I get $b = b$ which is true. 
I am stuck on the induction part. Here is am using the hint and saying that $s(n+1) = \displaystyle(n+1)\frac{a_0 + a_n}{2} + a_{n+1}$ 
Is this step right? Or do i also have to plug $n+1$ int the first $n$ of the equation? Where do I go from there?
 A: For the induction step, you say “Suppose the sum of $a_0, a_1,\ldots a_n$ is  $(n+1)\frac{a_0+a_n}{2}$. Then I need t0 show that the sum of $a_0, a_1,\ldots a_n, \color{red}{a_{n+1}}$ is $ (\color{red}{n+2})\frac{a_0+\color{red}{a_{n+1}}}{2}$.”  
So you are allowed to suppose the first part, that the sum of $a_0, a_1,\ldots a_n$ is  $(n+1)\frac{a_0+a_n}{2}$, and use that to show the second part, that the sum of $a_0, a_1,\ldots a_n, \color{red}{a_{n+1}}$ is $ (\color{red}{n+2})\frac{a_0+\color{red}{a_{n+1}}}{2}$.  If you can do that, you win.
Now you know that the sum of  $a_0, a_1,\ldots a_n, \color{red}{a_{n+1}}$ must be the same as the sum of  $a_0, a_1,\ldots a_n$, plus $\color{red}{a_{n+1}}$.  And you know from the "suppose" part that the sum of $a_0, a_1,\ldots a_n$ is  $(n+1)\frac{a_0+a_n}{2}$.  So that means that the sum of  $a_0, a_1,\ldots a_n, \color{red}{a_{n+1}}$ must be $(n+1)\frac{a_0+a_n}{2} + \color{red}{a_{n+1}}$. 
Can you show that this last expression is equal to $ (\color{red}{n+2})\frac{a_0+\color{red}{a_{n+1}}}{2}$?  If so, you win.  You'll need to use the fact that $a_i = b + i\cdot c$.
