$\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j 1 } } = \dots$

If $x_n = \frac{2n^2 + n + 1}{2n^2 - 3n + 2}$,then $\sum_{r=1}^n \left[ \left( \prod_{i=1}^r x_i \right) - 2\sum_{i=1}^r (2i - 1) \right]$ is equal to

Let $\langle a_n \rangle$ be a sequence such that $a_1+a_2+\ldots+a_n = \frac{n^2+3n}{(n+1)(n+2)}$. If $28 \sum_{k=1}^{10} \frac{1}{a_k} = p_1 p_2 p_3 \ldots p_m$,where $p_1, p_2, \ldots, p_m$ are the first $m$ prime numbers,then $m$ is equal to

The sum of the series $1 + (1 + 2) + (1 + 2 + 3) + \dots$ up to $n$ terms is:

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:

The sum of all positive integers $n$ for which $\frac{1^3+2^3+\ldots+(2n)^3}{1^2+2^2+\ldots+n^2}$ is also an integer is