How do I add it?

$\begin{aligned} f(x) & = 0 - 1 + 2 - 3 + \cdots + (-1)^x x \\ & = 0 - \left(1-2+3-\cdots + (-1)^{x+1} x \right) \\ & = 0 - p(x) \\ f(x) & = - p(x) \\ \implies f(x) + p(x+1) & = f(x) + p(x) \color{#3D99F6}{- (x+1)} \quad \quad \small \color{#3D99F6}{\text{Since }x \text{ is odd, }x+1 \text{ is even, hence add a '-' sign}} \\ & = f(x) - f(x) - x - 1 \\ & = \boxed{0-1-x} \end{aligned}$

f(x)\quad =\quad 0-1+2-3+4-5+6-7+8-9+.....+\begin{cases} -x,\quad if\quad x\quad is\quad odd \\ +x,\quad if\quad x\quad is\quad even \end{cases}\\ p(x)\quad =\quad 1-2+3-4+5-6+7-8+9-10+.....+\begin{cases} -x,\quad if\quad x\quad is\quad even \\ +x,\quad if\quad x\quad is\quad odd \end{cases}\\ \\

So, $f(x)$ can be written as

$f(x) = (-1)[1-2+3-4+5-6+7-8+9-10+ .... + \begin{cases} -x,\quad if\quad x\quad is\quad even \\ +x,\quad if\quad x\quad is\quad odd \end{cases}]$

That means, $f(x) = (-1) \times p(x)$

So, for $f(x)\quad and\quad p(x)$

$f(x) + p(x) = [(-1)\times p(x)] + [p(x)] = -p(x) + p(x) = p(x) - p(x) = 0$

Now, for $f(x) + p(x+1)$

$f(x) + p(x+1) = -p(x) + p(x+1) = -[1-2+3-4+ ... + x] + [1-2+3-4+ ... + x - (x + 1)] \\ \quad\quad\quad\quad = -[1-2+3-4+...+x] + [1-2+3-4+...+x] - (x+1)$

That yields,

$f(x) + p(x+1) = -(x+1) = -x - 1 = -1 - x = 0 - 1 - x$