A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of $$\mathrm{m}$$ is equal to:

For $$x \geqslant 0$$, the least value of $$\mathrm{K}$$, for which $$4^{1+x}+4^{1-x}, \frac{\mathrm{K}}{2}, 16^x+16^{-x}$$ are three consecutive terms of an A.P., is equal to :

If $$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$$ and $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=\mathrm{n}$$, then the point $$(\mathrm{m}, \mathrm{n})$$ lies on the line

The value of $$\frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots .+100^2 \times 101}$$ is