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Question

The value of $$\displaystyle\sum^{100}_{r=2}\dfrac{3^r(2-2r)}{(r+1)(r+2)}$$ is equal to?


A
123100100(101)
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B
323101101(102)
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C
323100100(101)
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D
None of these
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Solution

The correct option is B $$\dfrac{3}{2}-\dfrac{3^{101}}{101(102)}$$
$$ \displaystyle S= \sum_{r = 2}^{100} \dfrac{3^{r}(2-2r)}{r(r+1)(r+2)}$$
$$ \displaystyle S=\sum_{r = 2}^{100} \dfrac{3^{r}}{r(r+2)}-\dfrac{3^{r}+1}{(r+1)(r+2)}$$
$$ = \dfrac{3^{2}}{2(3)} - \dfrac{3^{3}}{3 \times 4}+ \dfrac{3^{3}}{3 \times 4} - \dfrac{3^{4}}{4\times 5}+\dfrac{3^{4}}{4\times 5}+...+\left ( \dfrac{3^{100}}{100 \times 101}-\dfrac{3^{101}}{101 \times 102} \right ) $$
$$ = \dfrac{3^{2}}{2(3)} - \dfrac{3^{101}}{101 \times 102 }$$
$$ S = \dfrac{3}{2} - \dfrac{3^{101}}{101 \times 102}$$

1183791_1069003_ans_2b04e9f83b4b45c5bf294b740c0fbbca.jpg

Mathematics

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