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Đáp án:
Bài 10:
a) $A=\dfrac{9}{20}$
b) $B=\dfrac{150}{31}$
c) $E=\left ( \dfrac{1}{2^{101}}-\dfrac{1}{2} \right ):\dfrac{-1}{2}$
d) $G=\left ( \dfrac{1}{2^{102}}-\dfrac{1}{2^{2}} \right ):\dfrac{-3}{4}$
Giải thích các bước giải:
Bài 10:
$a) A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{17.20}\\=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{17}-\dfrac{1}{20}\\=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\\b) B=\dfrac{5^{2}}{1.6}+\dfrac{5^{2}}{6.11}+\dfrac{5^{2}}{11.16}+...+\dfrac{5^{2}}{26.31}\\B=5\left ( \dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{26.31} \right )\\B=5\left ( 1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{26}-\dfrac{1}{31} \right )\\B=5\left ( 1-\dfrac{1}{31} \right )\\B=5.\dfrac{30}{31}=\dfrac{150}{31}\\c) E=\dfrac{1}{2}+\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+\dfrac{1}{2^{4}}+...+\dfrac{1}{2^{100}}\\\Leftrightarrow \dfrac{1}{2}E=\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+\dfrac{1}{2^{4}}+...+\dfrac{1}{2^{100}}+\dfrac{1}{2^{101}}\\\Rightarrow \dfrac{1}{2}E-E=\left (\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+\dfrac{1}{2^{4}}+...+\dfrac{1}{2^{100}}+\dfrac{1}{2^{101}} \right )-\left ( \dfrac{1}{2}+\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+\dfrac{1}{2^{4}}+...+\dfrac{1}{2^{100}} \right )\\\Leftrightarrow -\dfrac{1}{2}E=\dfrac{1}{2^{101}}-\dfrac{1}{2}\\\Leftrightarrow E=\left ( \dfrac{1}{2^{101}}-\dfrac{1}{2} \right ):\dfrac{-1}{2}\\d) G=\dfrac{1}{2^{2}}+\dfrac{1}{2^{4}}+\dfrac{1}{2^{6}}+...+\dfrac{1}{2^{100}}\\\Leftrightarrow \dfrac{1}{2^{2}}G=\dfrac{1}{2^{4}}+\dfrac{1}{2^{6}}+...+\dfrac{1}{2^{100}}+\dfrac{1}{2^{102}}\\\Rightarrow \dfrac{1}{2^{2}}G-G=\left ( \dfrac{1}{2^{4}}+\dfrac{1}{2^{6}}+...+\dfrac{1}{2^{100}}+\dfrac{1}{2^{102}} \right )-\left ( \dfrac{1}{2^{2}}+\dfrac{1}{2^{4}}+\dfrac{1}{2^{6}}+...+\dfrac{1}{2^{100}} \right )\\\Leftrightarrow -\dfrac{3}{4}G=\dfrac{1}{2^{102}}-\dfrac{1}{2^{2}}\\\Leftrightarrow G=\left ( \dfrac{1}{2^{102}}-\dfrac{1}{2^{2}} \right ):\dfrac{-3}{4}$
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