Initial value Problem with difference equation
Like a linear constant coefficient ODE the general solution can be written as the sum of a particular solution and the general solution of the homogeneous equation $x(n+1)+x(n)=0$.

Because the right hand side is a polynomial we look for a polynomial particular solution, and a bit of experimenting will show this to be a quadratic:$$x_p(n)=n^2/2-n/2$$

Also the general solution to the homogeneous equation is obviously: $x_h(n)=k(-1)^n$, and the solution to your problem is the sum $x_p(n)+x_h(n)$ with $k$ chosen to satisfy the initial condition.