If $4^3+8^3+12^3+\ldots$ up to $n$ terms $= k n^2(n+1)^2$ (for all $n \in N$),then $k=$

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:

$\frac{10001 \times 100 !}{2 \times 1 !+5 \times 2 !+10 \times 3 !+\ldots+10001 \times 100 !}=$

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = \frac{n}{n+1}$.

The sum of the first $n$ terms of the series $\frac{3}{5}+\frac{21}{25}+\frac{117}{125}+\ldots$ is

The value of the expression $1.(2 - \omega )(2 - {\omega ^2}) + 2.(3 - \omega )(3 - {\omega ^2}) + ....... + (n - 1).(n - \omega )(n - {\omega ^2}),$ where $\omega$ is an imaginary cube root of unity,is