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A=1/3*5 + 1/5*7 + ....+ 1/99*101
A=1/2(2/3*5 + 2/5*7 + ...+ 2/99*101)
A=1/2[(1/3-1/5)+(1/5-1/7)+...+(1/99-1/101)]
A=1/2(1/3-1/101)
A=1/2 * 98/303
A=49/303
\(A=\frac{1}{5}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(=\frac{1}{3X5}+\frac{1}{5X7}+\frac{1}{7X9}+\frac{1}{9X11}+...+\frac{1}{99X101}\)
\(2A=\frac{2}{3X5}+\frac{2}{5X7}+\frac{2}{7X9}+\frac{2}{9X11}+...+\frac{2}{99X101}\)
\(=\frac{5-3}{3X5}+\frac{7-5}{7X9}+\frac{9-7}{9X7}+\frac{11-9}{9X11}+...+\frac{101-99}{101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
Từ đó ta suy ra:\(A=\frac{98}{303}:2=\frac{49}{303}\)
2A=2/3.5+2/5.7+2/7.9+...+2/99.101=>2A=1/3-1/5+1/5-1/7+...+1/99-1/100=>2A=1/3-1/100=>2A=97/300=>A=97/600
B = 16/15 . 25/24 . ....... . 10000/9999
= 4.4/3.5 . 5.5/4.6 . ...... . 100.100/99.101
= 4.5. ..... .100/3.4. .... .99 . 4.5. ..... .100/5.6. ..... .101
= 100/3 . 4/101
= 400/303
Tk mk nha
A=\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
=\(\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
Quy đồng tính tiếp
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)
\(\Rightarrow\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(\Rightarrow\frac{1}{2}.\frac{98}{303}\)
\(\Rightarrow\frac{49}{303}\)
49/303,xin lỗi bạn mk làm biếng viết lời giải nếu cần nói mk nha
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