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câu 2: \(S=\frac{25^{28^{ }}+25^{24}+...+25^2+25^2+1}{25^{28}.25^2+25^{24}.25^4+...+25^2+1}\)
rút gọn ta được
\(S=\frac{1}{25^4+1}\)
Ta có : \(S=\frac{3}{2\cdot3}+\frac{3}{3\cdot6}+\frac{3}{4\cdot9}+...+\frac{3}{6039\cdot2014}\)
\(S=3\cdot\left(\frac{3}{6\cdot3}+\frac{3}{9\cdot6}+\frac{3}{12\cdot9}+...+\frac{3}{6039\cdot6042}\right)\)
\(S=3\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{6039}-\frac{1}{6042}\right)\)
\(S=3\cdot\left(\frac{1}{3}-\frac{1}{6042}\right)\)
\(S=3\cdot\frac{671}{2014}\)
\(S=\frac{2013}{2014}\)
a) Đặt A= \(\frac{1+2+2^2+2^3+...+2^{2009}}{1-2^{2010}}\)
Đặt S = 1 + 2 + 22 + 23 + ... + 22009
=> 2S = 2 + 22 + 23 + ... + 22010
=> 2S - S = (2 + 22 + 23 + ... + 22010) - (1 + 2 + 22 + 23 + .. + 22009)
=> S = 22010 - 1
=> S = - 1 - 22010
\(\Rightarrow A=\frac{-1-2^{2010}}{1-2^{2010}}=-1\)
Vậy \(\frac{1+2+2^2+2^3+...+2^{2009}}{1-2^{2010}}=-1\)
b) Đặt: \(A=\frac{1}{299.297}-\frac{1}{297.295}-\frac{1}{295.293}-...-\frac{1}{3.1}\)
\(\Rightarrow-2A=-\frac{2}{299.297}+\frac{2}{297.295}+\frac{2}{295.293}+...+\frac{2}{3.1}\)
\(\Rightarrow-2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{295.297}-\frac{2}{297.299}\)
\(\Rightarrow-2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{295}-\frac{1}{297}-\frac{1}{297.299}\)
\(\Rightarrow-2A=1-\frac{1}{297}-\frac{2}{88803}\)
\(\Rightarrow-2A=\frac{296}{297}-\frac{2}{88803}=\frac{88504}{88803}-\frac{2}{88803}=\frac{88502}{88803}\)
\(\Rightarrow A=\frac{88502}{88803}:\left(-2\right)=\frac{44251}{88803}\)
Vậy \(\frac{1}{299.297}-\frac{1}{297.295}-\frac{1}{295.293}-...-\frac{1}{3.1}=\frac{44251}{88803}\)
c) Đặt \(B=\frac{12}{1.3.5}+\frac{12}{3.5.7}+\frac{12}{5.7.9}+...+\frac{12}{25.27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{12}{25.27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{1}{1.3}-\frac{1}{27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow B=\frac{260}{783}.3=\frac{260}{261}\)
Vậy \(\frac{12}{1.3.5}+\frac{12}{3.5.7}+\frac{12}{5.7.9}+...+\frac{12}{25.27.29}=\frac{260}{261}\)
Duyệt mk nha!!!
\(S=\frac{3}{2.3}+\frac{3}{3.6}+...+\frac{3}{2014.6039}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2013.2014}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=1-\frac{1}{2014}=\frac{2013}{2014}\)
\(A=\frac{2.3+5^2.3^2}{5^2.1+2.3.3^2}=\frac{3\left(2+5^2.3\right)}{.......}\)
\(A=\frac{2.3+5^2.3^2}{5^2.1+2.3.3^2}=\frac{3\left(2+5^2.3\right)}{.......}\)