Use the Euler-Lagrange equation to find the equations of motion I have to use the Euler-Lagrange equation
$$
\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = 0
$$
to go from
$$
L = T - V = \sum_{n=1}^N \frac{m}{2} {\dot{q}_n}^2 - \sum_{n=1}^N \frac{K}{2} {\left( q_{n+1} - q_n \right)}^2
$$
to
$$
m\ddot{q}_n + \frac{K}{2} \left[ 2 \left(q_n - q_{n-1}\right) - 2 \left(q_{n+1} + - q_n\right) \right] = 0
$$
or
$$
m\ddot{q}_n = K \left( q_{n+1} + q_{n-1} - 2q_n \right)
$$
So I started with
$$
\frac{\partial L}{\partial \dot{q}} = \frac{m}{2} \sum_{n=1}^N \frac{\partial}{\partial \dot{q}} \left( {\dot{q}_n}^2 \right) = \sum_{n=1}^N m\dot{q}_n
$$
$$
\frac{d}{dt} \left( \sum_{n=1}^N m\dot{q}_n \right) = \sum_{n=1}^N m\ddot{q}_n
$$
Then on the second sum I'm missing something
$$
\frac{\partial L}{\partial q} = -\frac{K}{2} \sum_{n=1}^N \frac{\partial}{\partial q} {\left( q_{n+1} - q_n \right)}^2
$$
How does this get anywhere close to the 2nd term in the 3rd equation?
I figured that perhaps I should recalculate for every n, so rewriting the equations:
$$
\frac{\partial L}{\partial \dot{q}_n} = \frac{m}{2} \sum_{n=1}^N \frac{\partial}{\partial \dot{q}_n} \left( {\dot{q}_n}^2 \right) = \sum_{n=1}^N m\dot{q}_n
$$
$$
\frac{d}{dt} \left( \sum_{n=1}^N m\dot{q}_n \right) = \sum_{n=1}^N m\ddot{q}_n
$$
$$
\frac{\partial L}{\partial q_n} = -\frac{K}{2} \sum_{n=1}^N \frac{\partial}{\partial q_n} {\left( q_{n+1} - q_n \right)}^2 =
-\frac{K}{2} \sum_{n=1}^N 2{\left( q_{n+1} - q_n \right)}(-1)
$$
$$
\frac{\partial L}{\partial q_n} = \frac{K}{2} \sum_{n=1}^N 2{\left( q_{n+1} - q_n \right)}
$$
So for n+1
$$
\frac{\partial L}{\partial \dot{q}_{n+1}} = \frac{m}{2} \sum_{n=1}^N \frac{\partial}{\partial \dot{q}_{n+1}} \left( {\dot{q}_n}^2 \right) = 0
$$
$$
\frac{\partial L}{\partial q_{n+1}} = -\frac{K}{2} \sum_{n=1}^N \frac{\partial}{\partial q_{n+1}} {\left( q_{n+1} - q_n \right)}^2 =
-\frac{K}{2} \sum_{n=1}^N 2{\left( q_{n+1} - q_n \right)}
$$
Now there are 2 (coupled?) equations:
$$
m\ddot{q}_n - \frac{K}{2} 2{\left( q_{n+1} - q_n \right)} = 0
$$
$$
\frac{K}{2} 2{\left( q_{n+1} - q_n \right)} = 0
$$
Which result
$$
m\ddot{q}_n + K \left[ {\left( q_{n+1} - q_n \right) - \left( q_{n+1} - q_n \right)} \right] = 0
$$
If I can rewrite one of the parentheses n+1 -> n and n -> n-1, then I can get the desired result.
$$
m\ddot{q}_n + K \left[ {\left( q_{n} - q_{n-1} \right) - \left( q_{n+1} - q_n \right)} \right] = 0
$$
$$
m\ddot{q}_n + K {\left(- q_{n-1} - q_{n+1} + 2q_n \right)} = 0
$$
$$
m\ddot{q}_n = K {\left(q_{n-1} + q_{n+1} - 2q_n \right)}
$$
But this sounds like BS, so I have no idea what I'm doing. Enlighten me.
 A: We have
$$L= \frac{m}{2}\sum_{n~=~1}^N \dot{q}_n - \frac{K}{2}\sum_{n~=~1}^N(q_{n+1}-q_n)^2$$
Consider the equations of motion for the $k$-th particle:
$$ \frac{\partial L}{\partial q_k}-\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \dot{q}_k} = 0 $$.
The term of the velocity is easy:
$$ \frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \dot{q}_k} = \frac{\mathrm d}{\mathrm dt}(m \dot{q}_k)=m\ddot{q}_k$$
With the positions, notice that each coordinate appears in two terms, so
$$ \begin{align}\frac{\partial L}{\partial q_k} &= \frac{\partial}{\partial q_k}\left[-\frac{K}{2}(q_k-q_{k-1})^2-\frac{K}{2}(q_{k+1}-q_k)^2\right]
\\ &= -K(q_k-q_{k-1})+K(q_{k+1}-q_k)=K(q_{k+1}+q_{k-1}-2q_k)\end{align} $$ 
So, the eom for the $k$-th particle is
$$ m\ddot{q}_k = K(q_{k+1}+q_{k-1}-2q_k) $$
(obviously for $1<k<N$).
Some comments
Notice that I have intentionally changed the index for the derivation of the eom $k$ with respect to the dummy index $n$ that appears in the summatory. This is a very good practice when working with index notations that can help you to avoid mistakes when performing this calculations.
If you are somehow new to the index notations I recommend you to perform this computations explicitly for some given $N$ (let's say, $N=3$ or $4$), expanding the sums to actually view how it works and convince yourself that the method is right.
