If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

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....

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 $n = 1, 2, 3, 4, 5$.

Let $b_1, b_2, \dots, b_n$ be a geometric sequence such that $b_1 + b_2 = 1$ and $\sum\limits_{k = 1}^\infty b_k = 2$. Given that $b_2 < 0$,then the value of $b_1$ is:

The $n^{th}$ term of the series $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$ is

Find the sum to $n$ terms of the series $3 \times 8 + 6 \times 11 + 9 \times 14 + \dots$