For any integer $n \geq 1$,the sum $\sum_{k=1}^n k(k+2)$ is equal to

If the sum of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots + 2 \cdot (n-1)^2 + n^2$ (when $n$ is odd) is to be determined,given that for even $n$,the sum is $\frac{n(n+1)^2}{2}$,find the sum when $n$ is odd.

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:

If $n$ is odd or even,the sum of $n$ terms of the series $1 - 2 + 3 - 4 + 5 - 6 + \dots$ is given by:

Find the $20^{\text{th}}$ term in the following sequence whose $n^{\text{th}}$ term is $a_{n} = \frac{n(n-2)}{n+3}$.

In the sequence of sets $(1,2,3), (4,5,6), (7,8,9,10), \ldots$,the sum of elements in the $50^{th}$ set is