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Số số hạng của dãy trên là:
(103 - 1) : 2 + 1 = 52(số hạng)
Đúng đấy, ko hiểu thì hỏi nha.
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(\Rightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}...+\frac{2}{99.101}\)
\(\Rightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Rightarrow2A=\frac{1}{3}-\frac{1}{101}\)
\(\Rightarrow2A=\frac{101}{303}-\frac{3}{303}\)
\(\Rightarrow2A=\frac{98}{303}\)
\(\Rightarrow A=\frac{98}{303}:2=\frac{98}{303.2}=\frac{98}{606}=\frac{49}{303}\)
lên 820 điểm hỏi đáp nha
\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)
\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)
\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)
\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)
\(S=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)
\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{1}-\dfrac{1}{11}=\dfrac{11}{11}-\dfrac{1}{11}=\dfrac{10}{11}\)
a,\(\frac{2}{3.5}+\frac{2}{5.7}+.......+\frac{2}{11.13}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.............+\frac{1}{11}-\frac{1}{13}\)
=\(\frac{1}{3}-\frac{1}{13}\)
=\(\frac{10}{39}\)
b,Đặt A=\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.............+\frac{1}{27.28.29.30}\)
3A=\(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...........+\frac{3}{27.28.29.30}\)
3A=\(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+.............+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
3A=\(\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
3A=\(\frac{1}{6}-\frac{1}{24360}\)
3A=\(\frac{1353}{8120}\)
A=\(\frac{451}{8120}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{9\cdot11}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{9\cdot11}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{11-9}{9\cdot11}\right)\)
\(=\frac{1}{2}\left(\frac{3}{1\cdot3}-\frac{1}{1\cdot3}+\frac{5}{3\cdot5}-\frac{3}{3\cdot5}+...+\frac{7}{5\cdot7}-\frac{5}{5\cdot7}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(=\frac{1}{2}\cdot\frac{10}{11}\)
\(=\frac{10}{22}=\frac{5}{11}\)
Ta có :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(=\)\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\)\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\)\(\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(=\)\(\frac{1}{2}.\frac{10}{11}\)
\(=\)\(\frac{5}{11}\)
Bạn làm đúng òi
Chúc bạn học tốt ~
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}+\frac{13-11}{11.13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{5}{39}\)
\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{203.205}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{203.205}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{203}-\dfrac{1}{205}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{205}\right)\)
\(=\dfrac{1}{2}.\dfrac{202}{615}\)
\(=\dfrac{101}{615}\)
Chúc bạn học tốt!