$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$

Using $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x,$ we have

$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}=\left(\frac{1}{4} a\right)^{2}+\left(-\frac{1}{2} b\right)^{2}+(1)^{2}+2\left(\frac{1}{4} a\right)\left(-\frac{1}{2} b\right)+2\left(-\frac{1}{2} b\right)(1)+2(1)\left(\frac{1}{4} a\right)$

$\quad=\frac{1}{16} a^{2}+\frac{1}{4} b^{2}+1+\left[-\frac{1}{4} a b\right]+[-b]+\left[\frac{1}{2} a\right]$

$\quad=\frac{1}{16} a^{2}+\frac{1}{4} b^{2}+1-\frac{1}{4} a b-b+\frac{1}{2} a$