The largest perfect square that divides $2014^3 - 2013^3 + 2012^3 - 2011^3 + \ldots + 2^3 - 1^3$ is (in $^2$)

For $x \in \mathbb{R}$,let $[x]$ denote the greatest integer $\le x$. Find the sum of the series $\left[ -\frac{1}{3} \right] + \left[ -\frac{1}{3} - \frac{1}{100} \right] + \left[ -\frac{1}{3} - \frac{2}{100} \right] + \dots + \left[ -\frac{1}{3} - \frac{99}{100} \right]$.

The sum to $20$ terms of the series $2^2-3^2+4^2-5^2+6^2-\ldots$ is equal to $........$.

$\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$ is equal to

If $\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+1012}\right) - \left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots+\frac{1}{2024 \cdot 2023}\right) = \frac{1}{2024}$,then $\alpha$ is equal to-

Let $[\alpha]$ denote the greatest integer $\leq \alpha$. Then $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots +[\sqrt{120}]$ is equal to.