Sum of squares

Algebra Level 5

$1^2+2^2+3^2+\cdots+k^2 = \frac{(N-1)N(N+1)}{24},$ where $N = 2k+1$ .

Now consider the sum $(\tfrac12)^2+(1\tfrac12)^2+(2\tfrac12)^2+\cdots+(k-\tfrac12)^2.$ Can this sum also be written in the form $\frac{(N-1)N(N+1)}{24}\ ?$ If so, what is the value of $N$ ?