The sum to infinity of the progression $9 - 3 + 1 - \frac{1}{3} + \dots$ is

$\sum\limits_{r = 1}^n {\sum\limits_{m = 1}^r {m} } = \dots$

If the sum of the first $15$ terms of the series ${\left( {\frac{3}{4}} \right)^3} + {\left( {1\frac{1}{2}} \right)^3} + {\left( {2\frac{1}{4}} \right)^3} + {3^3} + {\left( {3\frac{3}{4}} \right)^3} + \dots$ is equal to $225\,k$,then $k$ 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

Let $S_n = \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2$. Then $S_n$ can take the value$(s)$:
$(A) 1056$
$(B) 1088$
$(C) 1120$
$(D) 1332$

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