Sum of squares

Algebra Level 5

12+22+32++k2=(N1)N(N+1)24,1^2+2^2+3^2+\cdots+k^2 = \frac{(N-1)N(N+1)}{24}, where N=2k+1N = 2k+1.

Now consider the sum (12)2+(112)2+(212)2++(k12)2.(\tfrac12)^2+(1\tfrac12)^2+(2\tfrac12)^2+\cdots+(k-\tfrac12)^2. Can this sum also be written in the form (N1)N(N+1)24 ?\frac{(N-1)N(N+1)}{24}\ ? If so, what is the value of NN?

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