Giải giúp mình gấp bài này với: S= \(-\frac{3^2}{4}\) - \(\frac{3^2}{28}\) - \(\frac{3^2}{70}\) - ...- \(\frac{3^2}{868}\) giải gấp giúp mình nha
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\(8\frac{4}{17}-\left(2\frac{5}{9}+3\frac{4}{17}\right)=\frac{140}{17}-\left(\frac{23}{9}+\frac{55}{17}\right)=\frac{140}{17}-\frac{886}{153}=\frac{22}{9}=2,444444444444\)
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\(S=\frac{3^2}{4}-\frac{3^2}{4.7}-\frac{3^2}{7.10}-...-\frac{3^2}{28.31}\)
\(S=\frac{3^2}{4}-\left(\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{28.31}\right)\)
\(S=\frac{9}{4}-3.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{28.31}\right)\)
\(S=\frac{9}{4}-3.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{28}-\frac{1}{31}\right)\)
\(S=\frac{9}{4}-3.\left(1-\frac{1}{31}\right)\)
\(S=\frac{9}{4}-3.\frac{30}{31}=\frac{9}{4}-\frac{90}{31}=\frac{-81}{124}\)
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\(\Leftrightarrow\left(\frac{3}{4}x-\frac{9}{16}\right)\left(\frac{1}{3}-\frac{3}{5}.\frac{1}{x}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{4}x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{4};\frac{9}{5}\right\}\)
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\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\cdot\cdot\frac{2016^2-1}{2016^2}=\frac{1.3}{2.3}\cdot\frac{2.4}{3.3}\cdot\cdot\cdot\cdot\frac{2015.2017}{2016.2016}\)
\(=\frac{\left(1.2.3....2015\right).\left(3.4....2016.2017\right)}{\left(2.3....2016\right)\left(2.3......2015.2016\right)}=\frac{2017}{2.2016}=\frac{2017}{4032}\)
S= - 32\(\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{868}\right)\)
S = - 32\(\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{28.31}\right)\)
S = - 3\(\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{28.31}\right)\)
S = -3\(\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{28}-\frac{1}{31}\right)\)
S = -3 \(\left(1-\frac{1}{31}\right)\)
S = -3\(.\frac{30}{31}\)
S = -90/31
1/3S=-(1/1*4+1/4*7+1/7*10+...+1/28*31)=-(1/1-1/4+1/4-1/7+1/7-1/10+...+1/28-1/31)=-(1/1-1/31)=-30/31
=>S=(-30/31):1/3=-90/31