What is the sum of $n$ terms of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$?

$2^2 + 4^2 + 6^2 + \dots + (2n)^2 = \dots$

Let $75 \ldots 57$ denote the $(r+2)$ digit number where the first and the last digits are $7$ and the remaining $r$ digits are $5$. Consider the sum $S = 77 + 757 + 7557 + \ldots + 75 \ldots 57$ (where the last term has $98$ digits). If $S = \frac{75 \ldots 57 + m}{n}$,where $m$ and $n$ are natural numbers less than $3000$,then the value of $m + n$ is:

If $S_n = 2 + 4 + 7 + 11 + \dots + n$ terms,then $t_n = \dots$

If $\sum_{k=1}^n \left( \sum_{m=1}^k m^2 \right) = an^4 + bn^3 + cn^2 + dn + e$,then which of the following is true?

The sum of the first $20$ terms of the series $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \dots$ is?