Find the sum of the series $6 + 66 + 666 + \dots$ up to $n$ terms.

The sequence of natural numbers is divided into groups as follows: $(1), (2, 3), (4, 5, 6), (7, 8, 9, 10), \dots$. Find the sum of the numbers in the $n^{th}$ group.

The sum $\frac{3 \times 1}{1^2} + \frac{5 \times (1^3 + 2^3)}{1^2 + 2^2} + \frac{7 \times (1^3 + 2^3 + 3^3)}{1^2 + 2^2 + 3^2} + \dots$ up to the $10^{th}$ term 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:

The value of $\frac{1 \times 2^2 + 2 \times 3^2 + \ldots + 100 \times 101^2}{1^2 \times 2 + 2^2 \times 3 + \ldots + 100^2 \times 101}$ is

Let $S_n = \sum_{k=1}^n (-1)^{k-1} \cdot k^2$ for $n \geq 1$. Given that $S_{2n} = -n(2n+1)$ for $n = 1, 2, 3, \ldots$,then $S_{77} =$