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 \(N\)?
by
Arjen Vreugdenhil