If sum_{n = 1}^5 frac{1}{n(n + 1)(n + 2)(n + | vedclass

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

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$\frac{55}{336}$

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(C) The general term of the given expression can be written using the method of differences as:$T_n = \frac{1}{3} \left[ \frac{1}{n(n + 1)(n + 2)} - \frac{1}{(n + 1)(n + 2)(n + 3)} \right]$Taking the summation from $n = 1$ to $5$ on both sides,we get a telescoping series:$\sum_{n = 1}^5 T_n = \frac{1}{3} \left[ \left( \frac{1}{1 \cdot 2 \cdot 3} - \frac{1}{2 \cdot 3 \cdot 4} \right) + \dots + \left( \frac{1}{5 \cdot 6 \cdot 7} - \frac{1}{6 \cdot 7 \cdot 8} \right) \right]$This simplifies to:$\sum_{n = 1}^5 T_n = \frac{1}{3} \left[ \frac{1}{6} - \frac{1}{6 \cdot 7 \cdot 8} \right] = \frac{k}{3}$$\Rightarrow \frac{1}{3} \left[ \frac{1}{6} - \frac{1}{336} \right] = \frac{k}{3}$$\Rightarrow \frac{1}{6} - \frac{1}{336} = k$$\Rightarrow k = \frac{56 - 1}{336} = \frac{55}{336}$

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

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