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

Find the sum of the series up to $10$ terms: $(3^3 - 2^3) + (5^3 - 4^3) + (7^3 - 6^3) + \dots$

The sum of the series $1 + 3x + 6x^2 + 10x^3 + \dots \infty$ is

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

Consider an infinite $G.P.$ with first term $a$ and common ratio $r$. Its sum is $4$ and the second term is $3/4$. 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....