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

The general term for the numerator is $$n(n+1)^2$$ and for the denominator is $$n^2(n+1)$$, evaluated from $$n=1$$ to $$100$$.

Let's expand both expressions:

• Numerator sum: $$\sum (n^3 + 2n^2 + n)$$
• Denominator sum: $$\sum (n^3 + n^2)$$

Next, apply the standard summation formulas for $$\sum n^3$$, $$\sum n^2$$, and $$\sum n$$:

• Numerator: $$\frac{n^2(n+1)^2}{4} + 2\left(\frac{n(n+1)(2n+1)}{6}\right) + \frac{n(n+1)}{2}$$
• Denominator: $$\frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6}$$

Now we can clearly pull out a common factor of $$\frac{n(n+1)}{2}$$ from every term in both the numerator and the denominator:

• Numerator: $$\frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + \frac{2(2n+1)}{3} + 1 \right] = \frac{n(n+1)}{2} \left[ \frac{3n^2 + 11n + 10}{6} \right]$$
• Denominator: $$\frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + \frac{2n+1}{3} \right] = \frac{n(n+1)}{2} \left[ \frac{3n^2 + 7n + 2}{6} \right]$$

Now, just factorize those resulting quadratics:

• Numerator quadratic: $$3n^2 + 11n + 10 = (3n+5)(n+2)$$
• Denominator quadratic: $$3n^2 + 7n + 2 = (3n+1)(n+2)$$

When you divide the numerator by the denominator, the initial $$\frac{n(n+1)}{2}$$ common factors, the $$6$$ in the denominators, and the $$(n+2)$$ terms all cancel out perfectly.

The entire series reduces to just: $$\frac{3n+5}{3n+1}$$

Since there are 100 terms, plug in $$n=100$$:

$$\frac{3(100) + 5}{3(100) + 1} = \frac{305}{301}$$