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\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)\)
= 1/2 .( 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + .......+ 1/2014.2015 - 1/2015.2016)
= 1/2 ( 1/2 - 1/2015.2016)
Tính tiếp p nhé.
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
B=2+1/1.2+2+1/2.3+........+2+1/9.10
B=2.9+1/1.2+1/2.3+........+1/9.10
B=18+9/10
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
= 1 / 1 - 1 / 2 + 1 / 2 - 1 / 3 + 1 / 3 - 1 / 4 + 1 / 4 - 1 / 5 + 1 / 5 - 1 / 6
Ta gạch các ps trùng.
Còn lại :
1 / 1 - 1 / 6 = 6 / 5
Áp dụng \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{49.50}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{306}{1225}\)
\(E=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
\(E=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)
\(E=2\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
\(E=2\left(\frac{1}{4x5}+\frac{1}{5x6}+...+\frac{1}{15x16}\right)\)
\(E=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(E=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(E=\frac{3}{8}\)
1/2E=1/20+1/30+1/42+...+1/240. =>1/2E=1/4*5+1/5*6+1/6*7+...+1/15*16. =>1/2E=1/4-1/5+1/5-1/6+1/6-1/7+...+1/15-1/16. =>1/2E=1/4-1/16=3/16. =>E=3/16:1/2=3/8. Câu b có vấn đề.
\(3C=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{27.28.29.30}\)
\(3C=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+...+\frac{30-27}{27.28.29.30}\)
\(3C=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}+\frac{1}{28.29.30}\)
\(3C=\frac{1}{1.2.3}-\frac{1}{28.29.30}\Rightarrow C=\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right):3\)