$1 + 3 + 7 + 15 + 31 + \dots$ to $n$ terms =

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 $f(0)=0, f(1)=1, f(2)=2$ and $f(x)=f(x-2)+f(x-3)$ for $x=3, 4, 5, \ldots$,then $f(10)=$

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.

For an odd integer $n \ge 1$,the value of $n^3 - (n-1)^3 + (n-2)^3 - (n-3)^3 + \dots + (-1)^{n-1} 1^3$ is:

The value of $\sum\limits_{r = 16}^{30} {(r + 2)(r - 3)}$ is equal to