If sumlimits_{n = 1}^5 {frac{1}{{nleft( {n + | vedclass

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(C) The general term $T_n$ can be expressed using the method of differences:$T_n = \frac{1}{3} \left[ \frac{1}{n(n+1)(n+2)} - \frac{1}{(n+1)(n+2)(n+3)

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$\frac{55}{336}$

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(C) The general term $T_n$ can be expressed using the method of differences:$T_n = \frac{1}{3} \left[ \frac{1}{n(n+1)(n+2)} - \frac{1}{(n+1)(n+2)(n+3)} \right]$Summing from $n=1$ to $5$:$\sum_{n=1}^5 T_n = \frac{1}{3} \left[ \left( \frac{1}{1 \cdot 2 \cdot 3} - \frac{1}{2 \cdot 3 \cdot 4} \right) + \dots + \left( \frac{1}{5 \cdot 6 \cdot 7} - \frac{1}{6 \cdot 7 \cdot 8} \right) \right]$This is a telescoping series:$\sum_{n=1}^5 T_n = \frac{1}{3} \left[ \frac{1}{6} - \frac{1}{6 \cdot 7 \cdot 8} \right] = \frac{1}{3} \left[ \frac{1}{6} - \frac{1}{336} \right]$$\frac{1}{3} \left[ \frac{56 - 1}{336} \right] = \frac{1}{3} \left( \frac{55}{336} \right) = \frac{k}{3}$Therefore,$k = \frac{55}{336}$.

If $\sum\limits_{n = 1}^5 {\frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}} = \frac{k}{3}} $,then $k$ is equal to

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