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

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(A) The $n$-th term of the series is given by ${T_n} = \frac{1 + 2 + \dots + n}{1^3 + 2^3 + \dots + n^3}$.Using the formulas for the sum of the first

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$199$

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(A) The $n$-th term of the series is given by ${T_n} = \frac{1 + 2 + \dots + n}{1^3 + 2^3 + \dots + n^3}$.Using the formulas for the sum of the first $n$ natural numbers and the sum of cubes of the first $n$ natural numbers:${T_n} = \frac{\frac{n(n+1)}{2}}{\left(\frac{n(n+1)}{2}ight)^2} = \frac{1}{\frac{n(n+1)}{2}} = \frac{2}{n(n+1)}$.Using partial fractions,${T_n} = 2\left(\frac{1}{n} - \frac{1}{n+1}ight)$.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 series: ${S_n} = 2\left( (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) ight) = 2\left(1 - \frac{1}{n+1}ight) = \frac{2n}{n+1}$.Given $100 S_n = n$,we have $100 \left(\frac{2n}{n+1}ight) = n$.Since $n 
eq 0$,we can divide by $n$: $\frac{200}{n+1} = 1$.Thus,$n+1 = 200$,which gives $n = 199$.

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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