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 $S_{77} =$

The sum to infinity of the following series $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots$ is:

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 odd numbers are divided as follows:
Row $1$: $1, 3$
Row $2$: $5, 7, 9, 11$
Row $3$: $13, 15, 17, 19, 21, 23$
Then the sum of the $n^{th}$ row is:

Let $u_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right]$,for $n = 0, 1, 2, ...$. Then which of the following is true?

The sum to infinity of the progression $9 - 3 + 1 - \frac{1}{3} + \dots$ is