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Đề sai rồi em, mẫu số đều là số lẻ thì 120 ko theo quy luật
Tổng các số đó là:
\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{399}\)
\(=\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+...+\dfrac{1}{19\times21}\)
\(=\dfrac{1}{2}\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{19\times21}\right)\)
\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{21}\right)\)
\(=\dfrac{1}{2}\times\dfrac{20}{21}\)
\(=\dfrac{10}{21}\)
A = \(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\) +...+
A = \(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\)+ \(\dfrac{1}{5.7}\) + \(\dfrac{1}{7.9}\)+...+
Xét dãy số 1; 3; 5; 7;...; Đây là dãy số cách đều với khoảng cách là
3 - 1 = 2
Số thứ 10 của dãy số trên là 2 x (10 - 1) + 1 = 19
Vậy tổng của mười phân số đầu tiên của tổng A là:
A = \(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + \(\dfrac{1}{7.9}\) +....+ \(\dfrac{1}{19.21}\)
A = \(\dfrac{2}{2}\).(\(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + \(\dfrac{1}{7.9}\) +...+ \(\dfrac{1}{19.21}\)
A = \(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + \(\dfrac{2}{7.9}\)+...+ \(\dfrac{2}{19.21}\))
A = \(\dfrac{1}{2}\). (\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + ...+ \(\dfrac{1}{19}\) - \(\dfrac{1}{21}\)
A = \(\dfrac{1}{2}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{21}\))
A = \(\dfrac{1}{2}\). \(\dfrac{20}{21}\)
A = \(\dfrac{10}{21}\)
a=78/35
b=22/12
c=1/1
d=40202090/4040090
e=1,24025667172...
f=871,82
ko biết đúng ko [0_0'] hihi
\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(=\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{11}-\dfrac{1}{13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{1}{3}+\dfrac{1}{25}\)
\(=\dfrac{28}{75}\)
\(\frac{4}{12}+\frac{4}{35}+\frac{4}{63}+\frac{4}{99}\)
\(=2.\left(\frac{2}{12}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)\)
\(=2.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=2.\left(\frac{11}{33}-\frac{3}{33}\right)\)
\(=2.\frac{8}{33}\)
\(=\frac{16}{33}\)
Tham khảo nhé~
A = 1/15 + 1/35 + 1/ 63 + 1/99 + ...+ 1/9999
A = 1/(3x5) + 1/(5x7) + 1/(7x9) + 1/(9x11) + ... + 1/(99 x 101)
Ax2 = 2/(3x5) + 2/(5x7) + 2/(7x9) + 2/(9x11) + ... + 2/(99 x 101)
Ax2 = 1/3 – 1/5 + 1/5 – 1/7 + 1/7 – 1/9 + 1/9 – 1/11 + ...+ 1/99 – 1/101
Ax2 = 1/3 – 1/101 = 98/303
A = 98/303 : 2
A = 49/303
\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{399}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(\frac{7}{21}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{6}{21}\)
\(=\frac{1}{7}\)
Chúc bạn học tốt !!!
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{399}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)
\(2A=\frac{1}{3}-\frac{1}{21}=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}=\frac{2}{7}\)
\(A=\frac{2}{7}:2=\frac{2}{7}.\frac{1}{2}=\frac{2}{14}=\frac{1}{7}\)
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)
\(\Rightarrow\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(\Rightarrow\frac{1}{2}.\frac{98}{303}\)
\(\Rightarrow\frac{49}{303}\)
\(S=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{120}\)
Bạn xem lại số 120 ở cuối nhé chứ như này là đề không ổn rồi
Em xem lại nhé, tổng S không có quy luật nên không tính được