$11^3 + 12^3 + \dots + 20^3$

Let $a, b, c, d$ be real numbers such that $\sum_{k=1}^n (a k^3+b k^2+c k+d)=n^4$,for every natural number $n$. Then,$|a|+|b|+|c|+|d|$ is equal to

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:

The numbers $a_n$ are defined by $a_0=1$ and $a_{n+1}=3n^2+n+a_n$ for $n \geq 0$. Then $a_n$ is equal to:

Find the sum to $n$ terms of the series $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \dots$

The odd numbers are divided as follows:
Row $1$: $1, 3$
Row $2$: $5, 7, 9, 11$
Row $3$: $13, 15, 17, 19, 21, 23$
Then the sum of the $n^{th}$ row is: