The value of frac{1 times 2^2 + 2 times 3^2 + | vedclass

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(B) Let the expression be $S = \frac{\sum_{r=1}^{n} r(r+1)^2}{\sum_{r=1}^{n} r^2(r+1)}$ where $n=100$. Numerator: $\sum_{r=1}^{n} (r^3 + 2r^2 + r) = \

Vedclass · Accepted Answer

$\frac{305}{301}$

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(B) Let the expression be $S = \frac{\sum_{r=1}^{n} r(r+1)^2}{\sum_{r=1}^{n} r^2(r+1)}$ where $n=100$. Numerator: $\sum_{r=1}^{n} (r^3 + 2r^2 + r) = \sum r^3 + 2\sum r^2 + \sum r = \frac{n^2(n+1)^2}{4} + \frac{2n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$. Denominator: $\sum_{r=1}^{n} (r^3 + r^2) = \sum r^3 + \sum r^2 = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6}$. Simplifying the ratio: $\frac{\frac{n(n+1)}{2} [\frac{n(n+1)}{2} + \frac{2(2n+1)}{3} + 1]}{\frac{n(n+1)}{2} [\frac{n(n+1)}{2} + \frac{2n+1}{3}]} = \frac{\frac{n^2+n}{2} + \frac{4n+2}{3} + 1}{\frac{n^2+n}{2} + \frac{2n+1}{3}}$. For $n=100$: $\frac{\frac{100 	imes 101}{2} + \frac{402}{3} + 1}{\frac{100 	imes 101}{2} + \frac{201}{3}} = \frac{5050 + 134 + 1}{5050 + 67} = \frac{5185}{5117}$. Dividing both by $17$,we get $\frac{305}{301}$.

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

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