The sum of all the products of the first $n$ natural numbers taken two at a time is

For a series $S = 1 - 2 + 3 - 4 + \dots$ up to $n$ terms,
Statement-$1$: The sum of the series is always dependent on the value of $n$,i.e.,whether it is even or odd.
Statement-$2$: The sum of the series is $-\frac{n}{2}$ when the value of $n$ is any even integer.

If $\sum\limits_{r=1}^\infty \frac{1}{(2r-1)^2} = \frac{\pi^2}{8}$,then $\sum\limits_{r=1}^\infty \frac{1}{r^2} = \dots$

The sum $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ up to $8$ terms, is:

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.

The sum of the series $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \dots$ to $n$ terms is: