nhân 100 với cả tử với mẫu sau đó phân tích mẫu ta có
100/1.99 +100/3.97+ ...+100/99.1 = (1+99)/1.99 +(3+97)/3.97 +...+ (1+99)/1.99
= 1/1.99+99/1.99+3/3.97+97/3.97+...+1/1.99+99/1.99
= 1/99 + 1/1 +1/3 +1/97+....+1/97+1/3+1/99+1/1
= 2.(1+1/3+1/5+1/7+.....1/97+1/99)
Do đó A =[100(1+1/3+1/5+..+1/99)] / 2(1+1/3+1/5+...1/99) = 50
nhân 100 với cả tử với mẫu sau đó phân tích mẫu ta có
100/1.99 +100/3.97+ ...+100/99.1 = (1+99)/1.99 +(3+97)/3.97 +...+ (1+99)/1.99
= 1/1.99+99/1.99+3/3.97+97/3.97+...+1/1.99+99/1.99
= 1/99 + 1/1 +1/3 +1/97+....+1/97+1/3+1/99+1/1
= 2.(1+1/3+1/5+1/7+.....1/97+1/99)
Do đó A =[100(1+1/3+1/5+..+1/99)] / 2(1+1/3+1/5+...1/99) = 50
\(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{99}\)
\(=\left(\frac{1}{99}+1\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+......+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\left(\frac{100}{1.99}\right)+\left(\frac{100}{3.97}\right)+.....+\left(\frac{100}{49.51}\right)\)
\(Ta\)\(thấy\)\(Biểu\)\(Thức\)\(Này\)\(gấp\)\(50\)\(biểu\)\(thức\)\(kia\)
\(\Rightarrow A=50\)
Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
