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1, A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\)
=4-\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)\)
= 4-\(\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)\)
=4-\(\left(1-\frac{1}{9}\right)\)
= 4-\(\frac{8}{9}\)
= \(\frac{7}{9}\)
Mình không hiểu quy luật của dãy số cho lắm! Xin lỗi.
(ngày mai bạn thi HSG môn Toán đúng không?)
\(M=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+...+1-\frac{1}{9999}\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)
\(M=50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(M=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(M=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{5050-100}{101}=\frac{4950}{101}\)
Đặt A=\(\frac{9}{10}+\frac{39}{40}+...+\frac{1119}{1120}\)
=>A=\(\frac{10-1}{10}+\frac{40-1}{40}+...+\frac{1120-1}{1120}\)
=>A=\(1-\frac{1}{10}+1-\frac{1}{40}+...+1-\frac{1}{1120}\)
=>A=\(11-\left(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{1120}\right)\)
Đặt B=\(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{1120}\)
=>3B=\(\frac{3}{10}+\frac{3}{40}+...+\frac{3}{1120}\)
=>3B=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{35}\)
=>3B=\(\frac{33}{70}\)
=>B=\(\frac{11}{70}\)
=>A=11-\(\frac{11}{70}\)
=>A=\(\frac{759}{70}\)
\(A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+...+\dfrac{9998}{9999}\\ =\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{9999}\right)\\ =\left(1+1+1+...+1\right)\left(\text{có 50 số 1}\right)-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\right)\\ =50\cdot1-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\right)\\ =50-\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =50-\left(1-\dfrac{1}{101}\right)\\ =50-1+\dfrac{1}{101}\\ =49+\dfrac{1}{101}\\ =\dfrac{4949+1}{101}\\ =\dfrac{4950}{101}\)