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\(N=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\)
\(1000N=1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+1\)
\(1000N=2\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)
\(N=\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)
\(\frac{M}{N}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]}=\frac{1}{\frac{1}{50}}=50\)
\(=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{1000\left(\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\right)}=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{\frac{1+999}{1.999}+\frac{3+997}{3.997}+...+\frac{997+3}{997.3}+\frac{999+1}{999.1}}\)
\(=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+\frac{1}{999}+1}=\frac{1000\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}=500\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)
\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\right)\)
\(A=2-\frac{1}{2^{2006}}\)
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\text{ }\)
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)
\(Q=0\)
Q=(1/99+12/999+123/999).(1/2-1/3-1/6) =(1/99+12/999+123/999).0 Q=0
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Làm thử thoi nhé :)
\(C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}}\)
\(\frac{1}{1000}C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1000}{1.999}+\frac{1000}{3.997}+...+\frac{1000}{997.3}+\frac{1000}{999.1}}\)
\(\frac{1}{1000}C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1+999}{1.999}+\frac{3+997}{3.997}+...+\frac{997+3}{997.3}+\frac{999+1}{999.1}}\)
\(\frac{1}{1000}C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{1.999}+\frac{999}{1.999}+\frac{3}{3.997}+\frac{997}{3.997}+...+\frac{997}{997.3}+\frac{3}{997.3}+\frac{999}{999.1}+\frac{1}{999.1}}\)
\(\frac{1}{1000}C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{999}+\frac{1}{1}+\frac{1}{997}+\frac{1}{3}+...+\frac{1}{3}+\frac{1}{997}+\frac{1}{1}+\frac{1}{999}}\)
\(\frac{1}{1000}C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)}\)
\(\frac{1}{1000}C=\frac{1}{2}\)
\(C=\frac{1}{2}.1000\)
\(C=500\)
Vậy \(C=500\)
Chúc bạn học tốt ~
1)Ta có:S=\(n_1^2+n_2^2+...+n_{10}^2\)=\(\left(n_1+n_2+...+n_{10}\right)^2-2.\left(n_1n_2+n_2n_3+.....+n_{10}.n_1\right)=2013^2-2.\left(n_1n_2+n_2n_3+.....+n_{10}.n_1\right)\)
Do 20132 chia 2 dư 1
\(2.\left(n_1n_2+n_2n_3+.....+n_{10}.n_1\right)\) chia hết cho 2
=>\(2013^2-2.\left(n_1n_2+n_2n_3+.....+n_{10}.n_1\right)-1\) chia hết cho 2
=>S-1 chia hết cho 2
Ác Mộng lam đủng rui. **** thui