The sum of the series $3.6 + 4.7 + 5.8 + \dots$ up to $(n - 2)$ terms is:

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

Find the sum to $n$ terms of the sequence,$8, 88, 888, 8888, \ldots$

If $2^3+4^3+6^3+\ldots+(2n)^3=h n^2(n+1)^2$,then $h$ is equal to

The value of $\sum\limits_{n = 0}^\infty {\frac{{{{(n + 1)}^2}}}{{{7^n}}}}$ is -