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1/3 + 1/15 + 1/35+ 1/63 +...... + 1/195
= 1/3 + 1/3x5 + 1/5 x7 + 1/7x9 + ....+1/13x15
= 1/3+1/3-1/5+1/5-1/7+1/7-1/9+....+1/13-1/15 ( vì +- nên rút gọn )
= 1/3+1/3-1/15
=3/5
=1/1.3+1/3.5+1/5.7+...+1/13.15
=1/2.2(1/1.3+1/3.5+1/5.7+...+1/13.15)
=1/2(2/1.3+2/3.5+2/5.7+...+2/13.15)
=1/2(1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15)
=1/2[(1-1/15)+(1/3-1/3)+(1/5-1/5)+...+(1/13-1/15)]
=1/2[(1-1/15)+0+...+0=1/2(1-1/15)=1/2.14/15=14/30=7/15
Viết 1 phân số vào dãy sau: \(\frac{1}{3};\frac{1}{15};\frac{1}{35};\frac{1}{63};...\)
Giúp mình với!
\(\frac{1}{3};\frac{1}{15};\frac{1}{35};\frac{1}{63}\)
Phân tích : \(\frac{1}{3}=\frac{1}{1\cdot3};\frac{1}{15}=\frac{1}{3\cdot5};\frac{1}{35}=\frac{1}{5\cdot7};\frac{1}{63}=\frac{1}{7\cdot9}\)
\(\Leftrightarrow n=\frac{1}{9\cdot11}=\frac{1}{99}\)
\(A=\frac{2}{3}+\frac{2}{15}+...+\frac{2}{143}\)
\(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{11\cdot13}\)
\(A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(A=1-\frac{1}{13}=\frac{12}{13}\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(=1-\frac{1}{13}\)
\(=\frac{12}{13}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2004\cdot2005}+\frac{1}{2005\cdot2006}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2004}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2006}\)
\(A=1-\frac{1}{2006}=\frac{2005}{2006}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(\Rightarrow A=1-\frac{1}{2006}\)
\(\Rightarrow A=\frac{2005}{2006}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}\)
\(=\frac{5}{12}\)
bn sẽ tinh theo kieeuranhaan 2 nha xin lỗi mik làm bi này rùi nhưng mik quên mik có sacks xem lại
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+....+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
