The Fibonacci sequence is defined by $a_1 = 1, a_2 = 1$ and $a_n = a_{n-1} + a_{n-2}$ for $n > 2$. Find $\frac{a_{n+1}}{a_n}$ for $n = 1, 2, 3, 4, 5$.

If the sum of the series $\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^2}-\frac{1}{2 \cdot 3}+\frac{1}{3^2}\right)+\left(\frac{1}{2^3}-\frac{1}{2^2 \cdot 3}+\frac{1}{2 \cdot 3^2}-\frac{1}{3^3}\right)+\left(\frac{1}{2^4}-\frac{1}{2^3 \cdot 3}+\frac{1}{2^2 \cdot 3^2}-\frac{1}{2 \cdot 3^3}+\frac{1}{3^4}\right)+\ldots$ is $\frac{\alpha}{\beta}$,where $\alpha$ and $\beta$ are co-prime,then $\alpha+3\beta$ is equal to....

What is the sum of $n$ terms of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$?

The value of ${(0.2)^{\log_{\sqrt{5}}\left( \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \infty \right)}}$ is:

Find the sum to $n$ terms of the series whose $n^{th}$ term is given by $(2n-1)^2$.

Find the sum of the following series up to $n$ terms:
$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots$