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

We need to find $$\frac{\sum_{r=1}^{100} r(r+1)^2}{\sum_{r=1}^{100} r^2(r+1)}$$.

Since $$r(r+1)^2 = r(r^2 + 2r + 1) = r^3 + 2r^2 + r$$, summing from $$r=1$$ to $$100$$ gives $$N = \sum_{r=1}^{100} r(r+1)^2 = \sum r^3 + 2\sum r^2 + \sum r\,. $$

Similarly, because $$r^2(r+1) = r^3 + r^2$$, we have $$D = \sum_{r=1}^{100} r^2(r+1) = \sum r^3 + \sum r^2\,. $$

Using standard formulas with $$n = 100$$, $$\sum_{r=1}^{n} r = \frac{n(n+1)}{2} = 5050\,,\qquad \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} = \frac{100\cdot101\cdot201}{6} = 338350\,,\qquad \sum_{r=1}^{n} r^3 = \Bigl[\frac{n(n+1)}{2}\Bigr]^2 = 5050^2 = 25502500\,. $$

Substituting these into the expressions for $$N$$ and $$D$$ gives $$N = 25502500 + 2\cdot338350 + 5050 = 25502500 + 676700 + 5050 = 26184250\,, $$ $$D = 25502500 + 338350 = 25840850\,. $$

Now, $$\frac{N}{D} = \frac{26184250}{25840850}\,. $$ Dividing both numerator and denominator by 50 yields $$\frac{523685}{516817}\,, $$ and further dividing by 1717 (since $$523685 = 305\times1717$$ and $$516817 = 301\times1717$$) gives $$\frac{N}{D} = \frac{305}{301}\,. $$

Therefore, the required value is $$\frac{305}{301}\,. $$