What is the sum of $n$ terms of the series $2 + 5 + 14 + 41 + \dots$?

If the sum of the first $15$ terms of the series ${\left( {\frac{3}{4}} \right)^3} + {\left( {1\frac{1}{2}} \right)^3} + {\left( {2\frac{1}{4}} \right)^3} + {3^3} + {\left( {3\frac{3}{4}} \right)^3} + \dots$ is equal to $225\,k$,then $k$ is equal to

The $9^{th}$ term of the sequence $27, 9, 5\frac{2}{5}, 3\frac{6}{7}, \dots$ is $.....$

Let $\alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \infty$ and $\beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots \infty$. Then the value of $(0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_{5}(\beta)}$ is equal to:

$t_1, t_2, t_3, \ldots, t_{n}$ are positive integers,$S_{n} = t_1 + t_2 + t_3 + \ldots + t_{n}$. Given $S_1 = 1^2, S_2 = 3^2, S_3 = 6^2, S_4 = 10^2, S_5 = 15^2$. Following this pattern,if $S_{10} = k^2$,then $k =$

The product $(32)(32)^{1/6}(32)^{1/36} \dots \infty$ is