Let S_n = frac{1}{1^3} + frac{1 + 2}{1^3 + 2^ | vedclass

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(A) The $n^{th}$ term of the series is given by $T_n = \frac{\sum_{k=1}^n k}{\sum_{k=1}^n k^3}$.Using the standard formulas $\sum_{k=1}^n k = \frac{n(

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$199$

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(A) The $n^{th}$ term of the series is given by $T_n = \frac{\sum_{k=1}^n k}{\sum_{k=1}^n k^3}$.Using the standard formulas $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}ight)^2$,we get:$T_n = \frac{\frac{n(n+1)}{2}}{\left(\frac{n(n+1)}{2}ight)^2} = \frac{2}{n(n+1)}$.We can write $T_n$ using partial fractions as $T_n = 2\left(\frac{1}{n} - \frac{1}{n+1}ight)$.Now,$S_n = \sum_{k=1}^n T_k = 2 \sum_{k=1}^n \left(\frac{1}{k} - \frac{1}{k+1}ight)$.This is a telescoping sum: $S_n = 2 \left( (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) ight) = 2 \left( 1 - \frac{1}{n+1} ight) = \frac{2n}{n+1}$.Given $100 S_n = n$,we substitute $S_n$:$100 \left( \frac{2n}{n+1} ight) = n$.Since $n 
eq 0$,we can divide by $n$:$\frac{200}{n+1} = 1 \implies n+1 = 200 \implies n = 199$.

Let $S_n = \frac{1}{1^3} + \frac{1 + 2}{1^3 + 2^3} + \frac{1 + 2 + 3}{1^3 + 2^3 + 3^3} + \dots + \frac{1 + 2 + \dots + n}{1^3 + 2^3 + \dots + n^3}$. If $100 S_n = n$,then $n$ is equal to:

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