What is the $20^{\text{th}}$ term of the sequence defined by $a_{n} = (n-1)(2-n)(3+n)$?

Find the sum of the following series up to $n$ terms:
$0.6 + 0.66 + 0.666 + \dots$

The sum of the infinite series $(\frac{1}{3}+\frac{4}{7})+(\frac{1}{3^{2}}+\frac{1}{3}\times\frac{4}{7}+\frac{4^{2}}{7^{2}})+(\frac{1}{3^{3}}+\frac{1}{3^{2}}\times\frac{4}{7}+\frac{1}{3}\times\frac{4^{2}}{7^{2}}+\frac{4^{3}}{7^{3}}) + \dots$ is equal to -

If $2^3+4^3+6^3+\ldots+(2n)^3 = h n^2(n+1)^2$,then $h$ is equal to

If $3 + \frac{1}{4} (3 + d) + \frac{1}{4^2} (3 + 2d) + \dots \infty = 8$,then the value of $d$ is:

$\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j 1 } } = \dots$