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

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Question

(C) The $n^{th}$ term $T_n$ is given by: $T_n = \frac{{\frac{n}{2} \cdot \frac{n+1}{2}}}{{\sum_{k=1}^{n} k^3}} = \frac{{\frac{n(n+1)}{4}}}{{{{\left( {

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The $n^{th}$ term $T_n$ is given by: $T_n = \frac{{\frac{n}{2} \cdot \frac{n+1}{2}}}{{\sum_{k=1}^{n} k^3}} = \frac{{\frac{n(n+1)}{4}}}{{{{\left( {\frac{{n(n + 1)}}{2}} \right)}^2}}}$ $= \frac{{\frac{{n(n + 1)}}{4}}}{{\frac{{{n^2}{{(n + 1)}^2}}}{4}}} = \frac{1}{{n(n + 1)}}$ Using partial fractions: $T_n = \frac{1}{n} - \frac{1}{{n + 1}}$ The sum $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{{k + 1}} \right)$ $S_n = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \dots + \left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)$ $S_n = 1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}}$.

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

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