$$\sum_{k=5}^{(n-1)-1\times \frac{1-(-1)^n}{2}} (n-k) \times \frac{1-(-1)^k}{2}$$
All n-values smaller or equal to the k-value are to be ignored.
if $$n=6$$
$$ans =1 $$
if $$n=7$$ $$ans=2$$
if $$n=8$$ $$ans=4$$
if $$n=9$$ $$ans=6$$
That's enough for the first one. Below is the second one.
$$\sum_{k=6}^{(n-1)-1 \times \frac{1+(-1)^n}{2}} \biggr((n-k)\times\frac{1+(-1)^k}{2}\biggl)\times2-1$$
if $$n=7$$ $$ans=1$$ if $$n=8$$ $$ans=3$$
if
$$n=9$$ $$ans=6$$
if $$n=10$$ $$ans=10$$
So, are the answers correct given the n-values? I am pretty confident in these, but I can't be too sure, because I'm quite new to precalculus, my calculator is a bit basic and I've never had a precalculus class. So, if any of you could check if these answers are correct, I'd appreciate it :)
Yes they are
First sum $$s_n=\sum _{k=5}^{(n-1)-\frac{1}{2} \left(1-(-1)^n\right)} \frac{1}{2} \left(1-(-1)^k\right) (n-k)$$
$$ \begin{array}{c|r} n & s_n \\ \hline 6 & 1 \\ 7 & 2 \\ 8 & 4 \\ 9 & 6 \\ 10 & 9 \\ 11 & 12 \\ 12 & 16 \\ 13 & 20 \\ 14 & 25 \\ 15 & 30 \\ \end{array} $$ general formula $$s_n=\frac{1}{8} \left(2 n^2-16 n+(-1)^n+31\right)$$
Second sum $$t_n=\sum _{k=6}^{(n-1)-\frac{1}{2} \left((-1)^n+1\right)} \left(\frac{2}{2} \left(\left((-1)^k+1\right) (n-k)\right)-1\right)$$ $$ \begin{array}{c|r} n & t_n \\ \hline 7 & 1 \\ 8 & 3 \\ 9 & 5 \\ 10 & 9 \\ 11 & 13 \\ 12 & 19 \\ 13 & 25 \\ 14 & 33 \\ 15 & 41 \\ \end{array} $$ general formula $$t_n=\frac{1}{4} \left(2 n^2-24 n+(-1)^n+75\right)$$