What is the sum of the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$?

Consider a sequence whose sum of first $n$ terms is given by $S_n = 4n^2 + 6n$,where $n \in N$. Then,the $15^{th}$ term $(T_{15})$ of this sequence is:

$\sum_{n=1}^5 n(n^2+n+1) = $

The sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is

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 $TRUE$?
$(A) T_{20}=1604$
$(B) \sum_{k=1}^{20} T_k=10510$
$(C) T_{30}=3454$
$(D) \sum_{k=1}^{30} T_k=35610$

If 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,what is the sum when $n$ is odd?