$2^2 + 4^2 + 6^2 + \dots + (2n)^2 = \dots$

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 sum to infinite terms of the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ is

If $\sum_{k=1}^{n} a_k = 6n^3$,then $\sum_{k=1}^{6} \left(\frac{a_{k+1}-a_k}{36}\right)^2$ is equal to . . . . . . .

Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots$. If $B - 2A = 100\lambda$,then $\lambda$ is equal to:

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = n \frac{n^{2}+5}{4}$.