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a) Giải
Đặt \(M=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\)
\(\Rightarrow A< A.M\)
hay \(A< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\right)\)
\(\Rightarrow A< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{98}{99}.\dfrac{99}{100}\)
\(\Leftrightarrow A< \dfrac{1.2.3.4.5.6...98.99}{2.3.4.5.6.7...99.100}\)
\(\Rightarrow A< \dfrac{1}{100}< \dfrac{1}{10}\)
Vậy \(A< \dfrac{1}{10}\)
Đáp án nè:
Đặt A=\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{99}}\)
3A=\(\dfrac{1}{1}-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
3A+A=\(\left(\dfrac{1}{1}-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
4A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)
4A bé hơn(sorry tớ không thấy dấu bé hơn)\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
Đặt B=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
3B=\(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)
4B=\(3-\dfrac{1}{3^{99}}\) bé hơn 3 \(\Rightarrow\)B bé hơn \(\dfrac{3}{4}\)
\(\Rightarrow\) 4A bé hơn\(\dfrac{3}{4}\Rightarrow\)A bé hơn \(\dfrac{3}{16}\)
Tick cho mình nha , ngồi đánh máy tính mỏi cả mắt lun
Chúc học tốt
Lời giải:
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(A+3A=1+\frac{1-2}{3}+\frac{-2+3}{3^2}+\frac{3-4}{3^3}+\frac{-4+5}{3^4}+...+\frac{99-100}{3^{99}}-\frac{100}{3^{100}}\)
\(4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-.....+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(4A=(1-\frac{1}{3})+(\frac{1}{3^2}-\frac{1}{3^3})+...+(\frac{1}{3^{98}}-\frac{1}{3^{99}})-\frac{100}{3^{100}}\)
\(4A=\frac{2}{3}+\frac{2}{3^3}+...+\frac{2}{3^{99}}-\frac{100}{3^{100}}\)
\(2A=\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{50}{3^{100}}\)
\(18A=3+\frac{1}{3}+...+\frac{1}{3^{97}}-\frac{450}{3^{100}}\)
\(\Rightarrow 18A-2A=3-\frac{1}{3^{99}}-\frac{450}{3^{100}}+\frac{50}{3^{100}}=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}\)
\(\Leftrightarrow 16A=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}<3\Rightarrow A< \frac{3}{16}\)
Đặt A=1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100
3A=1-2/3+3/3^2-4/3^3+...+99/3^98-100/3^99
3A+A=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99-100/3^100
<1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99
Đặt S=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99
3S=3-1+1/3-1/3^2+1/3^3-...-1/3^98
3S+S=3-1/3^99
S=(3-1/3^99) :4
S=3/4-1/4.3^99
\(\Rightarrow\)4A<3/4-1/4.3^99
\(\Rightarrow\)A<(3/4-1/4.3^99):4
\(\Rightarrow\)A<3/16-1/16.3^99<3/16
Vậy 1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100<3/16
\(A=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+....+\dfrac{\left(3^{99}+1\right)}{3^{99}}\)
\(A=\dfrac{4}{3}+\dfrac{10}{3^2}+\dfrac{28}{3^3}+...+\dfrac{\left(3^{99}+1\right)}{3^{99}}\)
\(A=\left(1+\dfrac{1}{3}\right)+\left(1+\dfrac{1}{3^2}\right)+\left(1+\dfrac{1}{3^3}\right)+...+\left(1+\dfrac{1}{3^{99}}\right)\)
\(A=\left(1+1+....+1\right)+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(A=99+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)
Gọi \(\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)là T
\(T=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)
\(3T=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)
\(3T-T=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)
\(2T=1-\dfrac{1}{3^{99}}\)
\(T=\left(1-\dfrac{1}{3^{99}}\right):2\)
\(T=\dfrac{1}{2}-\dfrac{1}{3^{99}\cdot2}\)
\(=>A=99+T=99+\dfrac{1}{2}-\dfrac{1}{3^{99}\cdot2}=99,5-\dfrac{1}{3^{99}\cdot2}< 100\)
Vậy A < 100
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