$\sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots+(2 k-1)}$ is equal to (in $.5$)

$t_1, t_2, t_3, \ldots, t_{n}$ are positive integers,$S_{n} = t_1 + t_2 + t_3 + \ldots + t_{n}$. Given $S_1 = 1^2, S_2 = 3^2, S_3 = 6^2, S_4 = 10^2, S_5 = 15^2$. Following this pattern,if $S_{10} = k^2$,then $k =$

If $4^3+8^3+12^3+\ldots$ up to $n$ terms $= k n^2(n+1)^2$ (for all $n \in N$),then $k=$

If $2.5+5.9+8.13+11.17+\ldots$ to $n$ terms $=an^3+bn^2+cn+d$,then $a-b+c-d=$

The sum to $20$ terms of the series $2^2-3^2+4^2-5^2+6^2-\ldots$ is equal to $........$.

If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots \infty = \frac{\pi^4}{90}$,$\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\ldots \infty = \alpha$,and $\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots \infty = \beta$,then $\frac{\alpha}{\beta}$ is equal to: