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\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
S=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{95.97}+\frac{1}{97.99}\)
S=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{99}\right)\)
S=\(\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
S=\(\frac{1}{2}.\frac{98}{99}\)
S=\(\frac{49}{99}\)
B2 : \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{114}+\frac{1}{196}+\frac{1}{256}+\frac{1}{324}\)
\(=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{18^2}\)
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{4^2}< \frac{1}{2\cdot4}\)
\(\frac{1}{6^2}< \frac{1}{4\cdot6}\)
...
\(\frac{1}{18}< \frac{1}{16\cdot18}\)
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{18^2}< \frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{16}-\frac{1}{18}\right)\)
\(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{18^2}< \frac{1}{2}< \frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{18}\right)\)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
Tự tính
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
A có tổng cộng 49 số hạng, nhóm 2 số hạng liên tiếp với nhau được:
\(A=\left(\frac{1}{1.3}-\frac{2}{3.5}\right)+\left(\frac{3}{5.7}-\frac{4}{7.9}\right)+...+\left(\frac{47}{93.95}-\frac{48}{95.97}\right)+\frac{49}{97.99}\)
\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{93.97}+\frac{49}{97.99}\)=> \(4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{93.97}+\frac{196}{97.99}=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{97}+\frac{196}{97.99}\)
=> \(4A=1-\frac{1}{97}+\frac{196}{97.99}=\frac{96}{97}+\frac{196}{97.99}=\frac{9700}{97.99}=\frac{100}{99}>1\)
\(4A>1=>A>\frac{1}{4}\)
A =(1/2 +1)×(1/3 +1)×(1/4 +1)×....×(1/99 +1)
=3/2x4/3x...............x100/99
=2-1/99
=197/99
A= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{100}{99}\)
A=\(\frac{\left(3\cdot4\cdot5\cdot....\cdot99\right)\cdot100}{2\cdot\left(3\cdot4\cdot5\cdot...\cdot99\right)}\)
A=\(\frac{100}{2}=50\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
=> \(\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)>\(\frac{32}{100}\)=32%
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{43.45}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{43.45}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{43}-\frac{1}{45}\)
\(=\frac{1}{3}-\frac{1}{45}=\frac{15}{45}-\frac{1}{45}=\frac{14}{45}\)
\(\Rightarrow A=\frac{14}{45}:2=\frac{14}{90}=\frac{7}{45}\)
Vậy \(A=\frac{7}{45}\).
Áp dụng công thức : \(\frac{1}{a}-\frac{1}{a+n}=\frac{n}{a\left(a+n\right)}\)
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{43\cdot45}\)
\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{43}-\frac{1}{45}\right)\)
\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{45}\right)\)
\(A=\frac{1}{2}\cdot\frac{14}{45}=\frac{7}{45}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(A=1-\frac{1}{101}\)
\(A=\frac{101}{101}-\frac{1}{101}\)
\(A=\frac{100}{101}\)
Chúc bạn học tốt !!!
A = 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/99 - 1/101
A = 1/1 - 1/101
A = 101/101 - 1/101
A = 100/101
\(A=\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.100}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}\)