frac{{frac{1}{2} cdot frac{2}{2}}}{{{1^3}}} + | vedclass

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(C) The $n^{th}$ term of the series is given by $T_n = \frac{{\frac{n(n + 1)}{2 \cdot 2}}}{{1^3 + 2^3 + 3^3 + \dots + n^3}}$.Using the formula for the

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$\frac{n}{{n + 1}}$

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(C) The $n^{th}$ term of the series is given by $T_n = \frac{{\frac{n(n + 1)}{2 \cdot 2}}}{{1^3 + 2^3 + 3^3 + \dots + n^3}}$.Using the formula for the sum of cubes,$\sum_{k=1}^n k^3 = \left( \frac{n(n+1)}{2} \right)^2$,we get:$T_n = \frac{{\frac{n(n + 1)}{4}}}{{\left( \frac{n(n + 1)}{2} \right)^2}} = \frac{{\frac{n(n + 1)}{4}}}{{\frac{n^2(n + 1)^2}{4}}} = \frac{1}{n(n + 1)}$.Using partial fractions,$T_n = \frac{1}{n} - \frac{1}{n + 1}$.The sum of $n$ terms $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k + 1} \right)$.Expanding the sum: $S_n = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{n} - \frac{1}{n + 1})$.All intermediate terms cancel out,leaving $S_n = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$.

$\frac{{\frac{1}{2} \cdot \frac{2}{2}}}{{{1^3}}} + \frac{{\frac{2}{2} \cdot \frac{3}{2}}}{{{1^3} + {2^3}}} + \frac{{\frac{3}{2} \cdot \frac{4}{2}}}{{{1^3} + {2^3} + {3^3}}} + \dots + n \text{ terms} =$

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