If the $n^{th}$ term of a sequence is $T_n = 2n - 1$,then the sum of $n$ terms $S_n = \dots$

For $\frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$ to exceed $1.01$,the maximum value of $n$ is

If the sum of $1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$ to $n$ terms is $S$,then $S$ is equal to

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....

$\sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots+(2 k-1)}$ is equal to (in $.5$)

The sum of the first $n$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots$ is $\frac{n(n + 1)^2}{2}$ when $n$ is even. When $n$ is odd,the sum is: