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Question: Evaluate \[ \sum_{k=1}^{100} \frac{1\cdot4}{2}+\frac{1\cdot4\cdot7}{2\cdot5}+\frac{1\cdot4\cdot7\cdo...

Evaluate

∑k=11001⋅42+1⋅4⋅72⋅5+1⋅4⋅7⋅102⋅5⋅8+…\sum_{k=1}^{100} \frac{1\cdot4}{2}+\frac{1\cdot4\cdot7}{2\cdot5}+\frac{1\cdot4\cdot7\cdot10}{2\cdot5\cdot8}+\dots
Answer

12(1⋅4⋅7⋅…⋅3012⋅5⋅8⋅…⋅299−3)\frac{1}{2}\left(\frac{1\cdot4\cdot7\cdot\ldots\cdot301}{2\cdot5\cdot8\cdot\ldots\cdot299}-3\right)

Explanation

Solution

Let Tk=1⋅4⋅7⋅⋯⋅(3k−2)2⋅5⋅8⋅⋯⋅(3k−1)T_k = \frac{1\cdot4\cdot7\cdot\dots\cdot(3k-2)}{2\cdot5\cdot8\cdot\dots\cdot(3k-1)}. We want to evaluate ∑k=1100Tk\sum_{k=1}^{100} T_k. The general term is Tk=1⋅4⋅7⋅⋯⋅(3k−2)2⋅5⋅8⋅⋯⋅(3k−1)T_k = \frac{1\cdot4\cdot7\cdot\dots\cdot(3k-2)}{2\cdot5\cdot8\cdot\dots\cdot(3k-1)}.

The sum of nn terms is given by the formula:

Sn=12(1⋅4⋅7⋅…⋅(3n−2)⋅(3n+1)2⋅5⋅8⋅…⋅(3n−1)−3)S_n = \frac{1}{2} \left( \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n-2) \cdot (3n+1)}{2 \cdot 5 \cdot 8 \cdot \ldots \cdot (3n-1)} - 3 \right).

Substituting n=100n=100:

∑k=1100Tk=12(1⋅4⋅7⋅…⋅(3⋅100−2)⋅(3⋅100+1)2⋅5⋅8⋅…⋅(3⋅100−1)−3)\sum_{k=1}^{100} T_k = \frac{1}{2} \left( \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3 \cdot 100-2) \cdot (3 \cdot 100+1)}{2 \cdot 5 \cdot 8 \cdot \ldots \cdot (3 \cdot 100-1)} - 3 \right)

∑k=1100Tk=12(1⋅4⋅7⋅…⋅298⋅3012⋅5⋅8⋅…⋅299−3)\sum_{k=1}^{100} T_k = \frac{1}{2} \left( \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot 298 \cdot 301}{2 \cdot 5 \cdot 8 \cdot \ldots \cdot 299} - 3 \right).

Therefore, the final answer is 12(1⋅4⋅7⋅…⋅3012⋅5⋅8⋅…⋅299−3)\frac{1}{2}\left(\frac{1\cdot4\cdot7\cdot\ldots\cdot301}{2\cdot5\cdot8\cdot\ldots\cdot299}-3\right).