Let $e_n$ be the sum of the squares of the first $n$ even positive integers, and let $o_n$ be the sum of the squares of the first $n$ odd positive integers; you have formulas for these. Let $s_n$ be the sum of the first $n$ terms of the target series, $1-4+9-16+-\dots$. Consider two cases, $n$ even and $n$ odd. (Note that I will be assuming throughout that $n$ is a positive integer.)
Suppose that $n$ is even, say $n=2m$. Then $\begin{align*} s_n&=s_{2m}\\ &=o_m-e_m\\ &=\frac{m(2m-1)(2m+1)}3-\frac{2m(m+1)(2m+1)}3\\ &=\frac{m(2m+1)}3\Big((2m-1)-2(m+1)\Big)\\ &=-m(2m+1)\;. \end{align*}$
Now suppose that $n$ is odd, say $n=2m+1$. Then $\begin{align*} s_n&=s_{2m+1}\\ &=o_{m+1}-e_m\\ &=\frac{(m+1)(2m+1)(2m+3)}3-\frac{2m(m+1)(2m+1)}3\\ &=(m+1)(2m+1)\;. \end{align*}$
Let’s run a quick spot-check. The first formula yields $s_4=-10$, which is correct, and the second yields $s_5=15$, which is also correct. All that remains is to combine them into a single nice expression, if possible. We have
$s_n=\begin{cases} -m(2m+1),&\text{if }n=2m\\ (m+1)(2m+1),&\text{if }n=2m+1\;. \end{cases}$
The algebraic sign is easily taken care of with a factor of $(-1)^{n+1}$, and $\lceil n/2\rceil=\begin{cases} m,&\text{if }n=2m\\ m+1,&\text{if }n=2m+1\;, \end{cases}$
so it only remains to handle the factor of $2m+1$. This can be done in many ways, but since we’re already using $\lceil n/2\rceil$, one easy way is to note that $2m+1=2\lceil n/2\rceil+(-1)^n$ no matter whether $n=2m$ or $n=2m+1$. Thus, we can finally write $s_n=(-1)^{n+1}\left\lceil\frac{n}2\right\rceil\left(2\left\lceil\frac{n}2\right\rceil+(-1)^n\right)\;,$ or, if you prefer, $s_n=2(-1)^{n+1}\left\lceil\frac{n}2\right\rceil^2-\left\lceil\frac{n}2\right\rceil\;.$