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\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{19}+\frac{1}{20}\right)\)\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)\)\(-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
\(=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)
Ta có:
\(\dfrac{11}{2}.\dfrac{12}{2}.\dfrac{13}{2}.....\dfrac{20}{2}\\ =\dfrac{11.12.13.....20}{2^{10}}\\ =\dfrac{\left(11.12.13.....20\right)\left(1.2.3.....10\right)}{2^{10}\left(1.2.3.....10\right)}\\ =\dfrac{1.2.3.4.....20}{2.4.6.8.....20}\\ =\dfrac{\left(1.3.5.7.....19\right)\left(2.4.6.....20\right)}{\left(2.4.6.....20\right)}\\ =1.3.5.7.....19\)
=> Đpcm
1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...+ 1/19 - 1/20
= ( 1 + 1/3 + 1/5 + ...+ 1/19 ) - ( 1/2 + 1/4 + ...+ 1/20 )
= ( 1 + 1/2 + 1/3 + 1/4 + ...+ 1/19 + 1/20 ) - 2 . ( 1/2 + 1/4 + ...+ 1/20 )
= ( 1 + 1/2 + 1/3 + ...+ 1/20 ) - ( 1 + 1/2 + ... + 1/10 )
= 1/11 + 1/12 + 1/13 + ...+ 1/20 ( Đpcm )
TK mk nha !!!
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{19}-\frac{1}{20}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{19}+\frac{1}{20}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}-1+\frac{1}{2}+....+\frac{1}{10}\)
\(=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\left(đpcm\right)\)
\(1.3.5...19=\frac{1.2.3.4...20}{2.4...20}=\frac{1.2.3.4...20}{2.1.2.2...2.10}=\frac{1.2.3.4....20}{2^{10}\left(1.2.3.4...10\right)}=\frac{11.12...20}{2^{20}}=\frac{11}{2}.\frac{12}{2}...\frac{20}{2}\)
sorry ae mình viết nhầm đề là 20/2 nha