TÍNH NHANH A=1/15+1/35+1/63+1/99+......+1/9999
AI NHANH TAY MÌNH TCK CHO
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=1/3*5+1/5*7+1/7*9+...+1/99*101
=1/3-1/5+1/5-1/7+...+1/99-1/101
=1/3-1/101
=98/303
\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(=\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{11}-\dfrac{1}{13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{1}{3}+\dfrac{1}{25}\)
\(=\dfrac{28}{75}\)
B = 1/3*5 + 1/5*7 + 1/7*9 + 1/9*11 + 1/11*13
= 1/2 * ( 2/3*5 + 2/5*7 + 2/7*9 + 2/9*11 + 2/11*13)
= 1/2 * ( 1/3 - 1/5 + 1/5 -1/7 + ...+ 1/11 - 1/13)
= 1/2 * ( 1/3 - 1/11)
= 1/2 * 8/33
= 4/33
\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(B=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(B=\frac{1}{2}.\frac{12}{39}\)
\(B=\frac{2}{13}\)
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)
\(\Rightarrow\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(\Rightarrow\frac{1}{2}.\frac{98}{303}\)
\(\Rightarrow\frac{49}{303}\)
49/303,xin lỗi bạn mk làm biếng viết lời giải nếu cần nói mk nha
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{101}\)
\(A=\left(\frac{1}{3}-\frac{1}{101}\right):2\)= 49/303
\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{99\times101}\)
\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+...+\frac{2}{99\times101}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\times\frac{98}{303}=\frac{49}{303}\)
Vậy A = 49/303.
A=\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+..+\frac{1}{9999}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+..+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+..+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{98}{303}=\frac{49}{303}\)
49/303 nha bạn
Kb với mình rồi mình giải kĩ cho
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