If $\frac{2+4+6+8+\dots+\text{upto } n \text{ terms}}{1+3+5+7+\dots+\text{upto } n \text{ terms}} = \frac{37}{36}$,then $n = $

If $1^2 + 2^2 + 3^2 + \dots + 2009^2 = (2009)(335)(4019)$ and $(1)(2009) + 2(2008) + 3(2007) + \dots + 2009(1) = (2009)(335)(x)$,then $x$ is equal to:

The value of $1^3 - 2^3 + 3^3 - \dots + 15^3$ is:

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

Let $a_1, a_2, a_3, \ldots$ be an arithmetic progression with $a_1=7$ and common difference $8$. Let $T_1, T_2, T_3, \ldots$ be such that $T_1=3$ and $T_{n+1}-T_n=a_n$ for $n \geq 1$. Then,which of the following is/are $TRUE$?
$(A) T_{20}=1604$
$(B) \sum_{k=1}^{20} T_k=10510$
$(C) T_{30}=3454$
$(D) \sum_{k=1}^{30} T_k=35610$

Find the sum of $n$ terms of the series $1 + 3 + 7 + 15 + 31 + \dots$