Proving $E(X)$ and $Var(X)$ for a random variable that has a discrete uniform distribution problem, what am I doing wrong I am having a problem proving $E(X)$ and $Var(X)$ for a random variable that has a discrete uniform distribution. The question and my current workings are below, can someone please indicate what I am doing wrong and if possible correct me?
Statement:  let $X=a,a+1,\dots,a+(s-1)$  be a random variable that has a discrete uniform distribution.
a. show that $E(X)=a+\frac{s-1}2$
method 1:
$$E(X)=\sum_{k=1}^{a+(s-1)}k\cdot P(X=x)=\sum_{k=1}^{a+(s-2)}k\cdot\left(a+\frac1{(s-2)}\right)\\
=a+\frac1{(s-2)}\sum_{k=1}^{(s-2)}k=a+\frac1{(s-2)}\left(\frac{(s-2)((s-2)+1)}{2}\right)\\
=a+\frac{(s-2)}{2}$$
method 2:
$$E(X)=\sum_{k=1}^{a+(s-1)}k\cdot P(X=x)=\sum_{k=1}^{a}k\cdot P(X=x)+\sum_{k=1}^{(s-1)}k\cdot P(X=x)\\
=\sum_{k=1}^a k\cdot\frac1a+\sum_{k=1}^{(s-1)} k\cdot\frac1{(s-1)}\\
=\frac1a\left(\frac{a(a+1)}{2}\right)+\frac1{(s-1)}\left(\frac{(s-1)((s-1)+1)}{2}\right)=\frac{a+1}{2}+\frac{s}{2}$$
b. show that $Var(X)=\frac{s^2-1}{12}$
$$Var(X)=E(X^2)-(E(X))^2$$
$$E(X^2)=\sum_{k=1}^{a+(s-1)}k^2\cdot P(X=x)=\sum_{k=1}^{a+(s-1)}k^2\cdot\frac{1}{a+(s-1)}=\frac{1}{a+(s-1)}\sum_{k=1}^{a+(s-1)}k^2\\
=\frac{1}{a+(s-1)}\left(\frac{(a+(s-1))(a+(s-1)+1)(2a+2(s-1)+1)}{6}\right)\\
=\frac{(a+s)(2a+2s+1}{6}=\frac{2a^2+4as-a+2s^2-s}{6}$$
Now
$$Var(X)=E(X^2)-(E(X))^2=\frac{2a^2+4as-a+2s^2-s}{6}-\left(a+\frac{s-1}{2}\right)^2\\
=\frac{2a^2+4as-a+2s^2-s}{6}-\left(a^2+a(s-1)+\frac{(s^2-2s-1)}{4}\right)\\
=\frac{2a^2+4as-a+2s^2-s}{6}-\left(\frac{4a^2+4as-4a+(s^2-2s-1)}{4}\right)\\
=\frac{4a^2+8as-2a+4s^2-2s-12a^2-12as+12a-3s^2+6s+3}{12}\\
=\frac{-8a^2-4as+10a+s^2+4s+3}{12}
$$

$$E(X^2) = \sum_{k=a}^{a+s-1} k^2P(X=k) = \frac 1{a+s-1}\sum_{k=a}^{a+s-1}k^2 = \frac{ (s-1)(6a^2 + 6as-6a + 2s^2 + 1 - 3s)}{6(a+s-1)}$$
$$Var(X) = E(X^2) - E(X)^2 = \frac{ (s-1)(6a^2 + 6as-6a + 2s^2 + 1 - 3s)}{6(a+s-1)} - a^2 - \frac {s^2 - 2s + 1}4 - a(s-1)$$
Now you could go ahead and simplify this, or using arguments similar to the one @BGM posted in the comments you can see that the above does not depend on $a$, so choose, say, $a = 0$ to find 
$$Var(X) = \frac{(s-1)(2s^2 - 3s + 1)}{6(s-1)} - \frac{s^2 - 2s + 1}{4} = \frac {s^2}3 - \frac s2 + \frac 16 - \frac{s^2}4 + \frac s2 - \frac 14 = \frac {s^2}{12}-\frac 1{12} $$
