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:

The sum of an infinite geometric series is $3$. $A$ series,which is formed by squaring its terms,also has a sum of $3$. The first series is

The sum of the first $n$ terms of the series $\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots$ is

If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots \infty = \frac{\pi^4}{90}$,$\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\ldots \infty = \alpha$,and $\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots \infty = \beta$,then $\frac{\alpha}{\beta}$ is equal to:

If $1 + \sin \theta + \sin^2 \theta + \dots \text{ to } \infty = 4 + 2\sqrt{3}$,where $0 < \theta < \pi$ and $\theta \neq \frac{\pi}{2}$,then $\theta = $

If the sum of the first $n$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots$ is $\frac{n(n+1)^2}{2}$ when $n$ is even,what is the sum when $n$ is odd?