Variance of the dicrete uniform distribution.

Can anybody help,Show that the Variance of the dicrete uniform distribution f(x)=

{1/n, x=1,2,3,4,.....,n}
{0 , elsewhere.}

is given by variance=[n^2-1]/12

Just resort to the definition and a couple of well-known results.

E[x] = (1/n)[1 + 2 + 3 + ... + n] = (1/n)(n(n+1)/2) = (n+1)/2

E[x^2] = (1/n)[1 + 4 + 9 + ... + n^2] = (1/n)(n(n+1)(2n+1)/6) = ((n+1)(2n+1)/6)

Var(x) = E[x^2] - (E[x])^2

Thanks very much tkhunny I understood u.