The sum of the following series $1 + 6 + \frac{9(1^2 + 2^2 + 3^2)}{7} + \frac{12(1^2 + 2^2 + 3^2 + 4^2)}{9} + \frac{15(1^2 + 2^2 + ... + 5^2)}{11} + ...$ up to $15$ terms 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 $f(1)=3$, and $f(n+1)-f(n)=3(4^n-1)$, then for all $n \in N$, $f(n)=$

$11^2 + 12^2 + 13^2 + \dots + 20^2 = ?$

The sum to infinity of the following series $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots$ is

The sum of the first $n$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots$ is $\frac{n(n + 1)^2}{2}$ when $n$ is even. When $n$ is odd,the sum is: