Solve the recurrence of the alternating sum $R_n=R_{n-1}+(-1)^{n}(n+1)^{2}$ I have been trying to solve this recurrence for a few hours, but I haven't been able to find the solution yet:
$R_0=1$
$R_n=R_{n-1}+(-1)^{n}*(n+1)^{2}$.
I have been trying to substitute $T_n=(-1)^{n}*R_n$ and then solving for $T_n$ and got the sum:
$1+3+...+\frac{n(n+1)}{2}$
but this sum $(-1)^{n}\frac{n(n+1)(n+2)}{6}$ didn't give the generalized form that would give the terms of the recurrence.
Additionally, I have been trying to substitute $n=2*a$:
$\sum_{k=1}^a (-1)^{2k}*(2k+1)^{2}+ \sum_{k=1}^a (-1)^{2k-1}*(2k)^2$
and I got $\frac{n*(n+3)}{2}$, but it doesn't seem to be right either. 
Please help me and if you can figure out, please tell me what I did wrong. 
Edited: I have added the initial values.

@hypergeometric's solution is my favorite, but another (very general) way of solving this recurrence is by a so-called generating function. Let us define $f(x) = \sum_{n=0}^\infty R_n x^n$. Then,
$$ R_n = R_{n-1} + (-1)^n(n+1)^2 \implies R_nx^n = x R_{n-1} x^{n-1} + (-1)^n (n+1)^2 x^n$$
Summing both sides from $n=1$ to $\infty$, we obtain
$$ f(x)-1 = xf(x) + \sum_{n=1}^\infty (-1)^n (n+1)^2 x^n $$
We can now evaluate the remaining sum by noting
$$ \frac{d^2}{dx^2} \frac{1}{1-x} = \sum_{n=2}^\infty n(n-1)x^{n-2} = \sum_{n=0}^\infty (n+2)(n+1) x^n = \sum_{n=0}^\infty (n+1)^2 x^n + \sum_{n=0}^\infty (n+1)x^n$$
Thus (changing the limits of summation from $n=0$ to $n=1$),
$$ \frac{2}{(1-x)^3} - \frac{1}{(1-x)^2} - 1 = \sum_{n=1}^\infty (n+1)^2 x^n $$
This (substituting $-x$ for $x$) gives us
$$ (1-x)f(x) = \frac{2}{(1+x)^3} - \frac{1}{(1+x)^2} \implies f(x) = \frac{1}{(1+x)^3} = \sum_{n=0}^\infty \underbrace{\frac{(-1)^n}{2}(n+1)(n+2)}_{R_n}x^n.$$
The trick was that we found $f(x)$ in closed form, then re-expanded as a series, since it was of a well-known form. The coefficients of that series are, by construction, $R_n$.
