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

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 $n^{th}$ term of a series is $n(n + 1)$,then the sum of its $n$ terms is......

The sum of the first $10$ terms of the series $9+99+999+\ldots$ is

Let $a_{n}$ be the $n^{\text{th}}$ term of the series $5+8+14+23+35+50+\ldots$ and $S_{n}=\sum_{k=1}^{n} a_{k}$. Then $S_{30}-a_{40}$ is equal to

The sum of $1^3 + 2^3 + 3^3 + 4^3 + \dots + 15^3$ is