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\(S=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+.......+\frac{61}{\left(30.31\right)^2}\)
\(=\frac{1}{1^2.2^2}+\frac{1}{2^2.3^2}+....+\frac{1}{30^2.31^2}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{30}-\frac{1}{31}\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-......-\left(\frac{1}{30}-\frac{1}{30}\right)-\frac{1}{31}\)
\(=1-\frac{1}{31}\\ =\frac{31}{31}-\frac{1}{31}=\frac{30}{31}\)
no mình nha
Với \(n\ge1\)thì \(\frac{2n+1}{n^2\left(n+1\right)^2}=\frac{n^2+2n+1-n^2}{n^2\left(n+1\right)^2}=\frac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\frac{\left(n+1\right)^2}{n^2\left(n+1\right)^2}-\frac{n^2}{n^2\left(n+1\right)^2}\)
Do đó \(S=\frac{3}{\left(1\cdot2\right)^2}+\frac{5}{\left(2\cdot3\right)^2}+...+\frac{4017}{\left(2008\cdot2009\right)^2}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{2008^2}-\frac{1}{2009^2}\)
\(=1-\frac{1}{2009^2}\)
sao bạn hôm đăng bài lớp 8 hôm thì đăng bài lớp 6 vậy
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
\(\frac{27^3.4^5}{6^8}:\left(\frac{5^5.2^4}{10^4}:\frac{6^4}{2^6.3^4}\right)\)
\(=\frac{3^9.2^{10}}{6^8}:\left(5:\frac{1}{2^2}\right)\)\(=3.2^2:20=\frac{12}{20}=\frac{3}{5}\)
\(\frac{27^3.4^5}{6^8}:\left(\frac{5^5.2^4}{10^4}:\frac{6^4}{2^6.3^4}\right)\)
\(=\frac{\left(3^3\right)^3.\left(2^2\right)^5}{\left(3.2\right)^8}:\left(\frac{5^5.2^4}{\left(5.2\right)^4}:\frac{\left(2.3\right)^4}{2^6.3^4}\right)\)
\(=\frac{3^9.2^{10}}{3^8.2^8}:\left(\frac{5^5.2^4}{5^4.2^4}:\frac{2^4.3^4}{2^6.3^4}\right)\)
\(=3.2^2:\left(5:\frac{1}{2^2}\right)\)
\(=3.4:\left(5.4\right)\)
\(=12:20\)
\(=\frac{12}{20}=\frac{3}{5}\)
\(\left(1\cdot2\right)^{-1}+\left(2\cdot3\right)^{-1}+\cdot\cdot\cdot+\left(9\cdot10\right)^{-1}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(S=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{61}{\left(30.31\right)^2}\)
\(S=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{61}{30^2.31^2}\)
\(S=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{61}{900.961}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{900}-\frac{1}{961}\)
\(S=1-\frac{1}{961}\)
\(S=\frac{960}{961}\)