$\sum_{n=1}^5 n(n^2+n+1) = $

The sum of an infinite number of terms in a $G.P.$ is $20$ and the sum of their squares is $100$. The common ratio of the $G.P.$ is:

The sum of $(1^2-1+1)(1!) + (2^2-2+1)(2!) + \ldots + (n^2-n+1)(n!)$ is

The sum $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ up to $8$ terms, 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:

Find the sum of the series $2 + 4 + 7 + 11 + 16 + \dots$ up to $n$ terms.