$$\begin{aligned} & \text { Given, } \mathrm{S}_n=\sum_{k=1}^{4 n}(-1)^{\frac{k(k+1)}{2}} \cdot \mathrm{K}^2 \\ & \Rightarrow \mathrm{S}_n=-1^2-2^2+3^2+4^2-5^2-6^2+7^2 +8^2-9^2-10^2+11^2+12^2 \ldots \ldots . 4 n \text { terms } \end{aligned}$$

$$\begin{aligned} \Rightarrow \mathrm{S}_n= & -\left[1^2+5^2+9^2+\ldots . n \text { terms }\right] \\ & -\left[2^2+6^2+10^2+\ldots n \text { terms }\right] \\ & +\left[3^2+7^2+11^2+\ldots n \text { terms }\right] \\ & +\left[4^2+8^2+12^2+\ldots n \text { terms }\right] \end{aligned}$$

$$\Rightarrow \mathrm{S}_n=-\sum_\limits{r=1}^n(4 r-3)^2-\sum_\limits{r=1}^n(4 r-2)^2 +\sum_\limits{r=1}^n(4 r-1)^2+\sum_\limits{r=1}^n(4 r)^2$$

$$\begin{aligned} & \Rightarrow \mathrm{S}_n=\sum_{r=1}^n\left((4 r)^2+(4 r-1)^2-(4 r-2)^2-(4 r-3)^2\right) \\ & \Rightarrow \mathrm{S}_n=\sum_{r=1}^n(32 r-12) \\ & \Rightarrow \mathrm{S}_n=32 \sum_{r=1}^n r-\sum_{r=1}^n 12 \\ & \Rightarrow \mathrm{S}_n=32 \cdot \frac{n(n+1)}{2}-12 n \\ & \Rightarrow \mathrm{S}_n=16 n^2+16 n-12 n \\ & \Rightarrow \mathrm{S}_n=4 n(4 n+1) \end{aligned}$$

If $$n=9$$, then $$\mathrm{S}_9=1332$$

If $$n=8$$, then $$\mathrm{S}_8=1056$$

Hints :

(i) Recall $$\sum_\limits{\mathrm{K}=1}^{4 n}(-1)^{\frac{k(k+1)}{2}} \cdot \mathrm{K}^2=-\sum_\limits{r=1}^n(4 r-3)^2-\sum_\limits{r=1}^n(4 r-2)^2+\sum_\limits{r=1}^n(4 r-1)^2+\sum_\limits{r=1}^n(4 r)^2$$

(ii) $$\sum_\limits{r=1}^n a \cdot f(r)+b \cdot g(r)-c \cdot h(r) = \sum_\limits{r=1}^n a \cdot f(r)+\sum_\limits{r=1}^n b \cdot g(r)-\sum_\limits{r=1}^n c \cdot h(r)$$

$$=a \sum_\limits{r=1}^n f(r)+b \sum_\limits{r=1}^n g(r)-c \sum_\limits{r=1}^n h(r)$$

(iii) $$\sum_\limits{r=1}^n r=\frac{n(n+1)}{2}, \sum_\limits{r=1}^n a=a r$$ where is a constant.