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?
 A: You can directly plug in $c=(u+d)/2$ and expand out the algebra as follows:
$$\frac{1+u}{2}(1+u+d)+\frac{1+d}{2}\frac{2+u+d}{2}=\left(\frac{2+u+d}{2}\right)^2+\frac{1+u}{2}\frac{u+d}{2}$$
$$2(1+u)(1+u+d)+(1+d)(2+u+d)=(2+u+d)^2+(1+u)(u+d)$$
$$4+5u+5d+3ud+2u^2+d^2=4+5u+5d+3ud+2u^2+d^2$$
(some steps purposely omitted)
and thus the two sides are equal, regardless of restrictions on $u$ and $d$.
A: $$
\frac{1 + u}{2}(1 + 2c) + \frac{1+d}{2}(1 + c) = \frac{1 + u}{2}(1 + c) + \frac{1 + u}{2}c + \frac{1+d}{2}(1 + c)
$$
$$
=(1 + c)\biggl(\frac{1 + u}{2} + \frac{1 + d}{2} \biggr) + \frac{1 + u}{2}c = (1 + c)\biggl(1 + \frac{u + d}{2}\biggr) + \frac{1 + u}{2}c
$$
$$
= (1 + c)^2 + \frac{1 + u}{2}c
$$
A: 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 $2c-d=u$:
$$\frac{1}{2}(c+uc) + \frac{1}{2}(2+4c-d+(2c-d)c+dc+d)$$
$$\frac{1}{2}(c+uc)+\frac{1}{2}(2+4c+2c^2)$$
$$\frac{1}{2}(c+uc)+(c+1)^2$$
