Equal allocation minimises a separable sum of squares

The dynamic programming approach to an optimization problem gives the recurrence equation for optimal solution as $f_k(b_k)=\min_{0<x\le b}\{x^2+f_{k-1}(b_{k-1})\}$ for $k>1$, with a base case reducing the last stage to using all remaining resource. Then, $f_3(15)=$