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Answers ( 1 )
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    A= \(-\dfrac{1}{3}+\dfrac{1}{3^2}-...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{101}}\)

    =>3A=\(-1+\dfrac{1}{3}-...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)

    =>4A=\(-1-\dfrac{1}{3^{101}}\) =>A=\(\dfrac{-1-\dfrac{1}{3^{101}}}{4}\)

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