The mean of the squares of the first $n$ natural numbers is

If $n$ is the smallest natural number such that $n+2n+3n+\ldots+99n$ is a perfect square,then the number of digits of $n^2$ is

$11^2 + 12^2 + 13^2 + \dots + 20^2 = ?$

If $\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots + \infty = \frac{\pi^4}{90}$,then the value of $\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \dots + \infty$ is

$2 + 4 + 7 + 11 + 16 + \dots$ to $n$ terms =

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....