Easy algebra question: show equality? Let $c:=(u+d)/2$ and $-1<d<0<u$.
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Can you show the steps to arrive from lhs to rhs?

Expanding the left hand side:
$$\frac{1}{2}(1+u+2c+2uc)+\frac{1}{2}(1+d+dc+c)$$
$$\frac{1}{2}(2+u+c+ 2c + uc + uc +dc +d)$$
$$\frac{1}{2}(c+uc) + \frac{1}{2}(2+u+2c+uc+ dc+d)$$
Taking into account that $2c-d=u$:
$$\frac{1}{2}(c+uc) + \frac{1}{2}(2+4c-d+(2c-d)c+dc+d)$$
$$\frac{1}{2}(c+uc)+\frac{1}{2}(2+4c+2c^2)$$
$$\frac{1}{2}(c+uc)+(c+1)^2$$
