Chứng minh rằng:
\(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
chứng minh rằng:
\(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
Chứng minh rằng: A=\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
Chứng minh rằng : \(B=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
\(B=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
\(3B=1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+...+\dfrac{100}{3^{99}}\)
\(3B-B=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{100}{3^{100}}\right)\)
\(2B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6B=3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(6B-2B=\left(3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)\)
\(4B=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)
\(4B=3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+\dfrac{100}{3^{100}}\)
\(4B=3-\dfrac{203}{3^{100}}< 3\)
\(\Rightarrow C< \dfrac{3}{4}\left(đpcm\right)\)
nhân 3 trừ đi sau đó xét cái sau sẽ thấy B<3/4
Lười lắm
\(B=\dfrac{1}{3}+\dfrac{2}{3^2}+...+\dfrac{100}{3^{100}}\\ \Rightarrow3B=1+\dfrac{2}{3}+....+\dfrac{100}{3^{99}}\\ \Rightarrow2B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\\ A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^{99}}\\ \Rightarrow3A=3+1+..+\dfrac{1}{3^{98}}\\ \Rightarrow2.A=3-\dfrac{1}{3^{99}}< 3\\ \Rightarrow A< \dfrac{3}{2}\\ \Rightarrow2B=A-\dfrac{100}{3^{100}}< \dfrac{3}{2}\\ \Rightarrow B< \dfrac{3}{4}\)
Chứng minh rằng: \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
Chứng minh rằng:
\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
Lời giải:
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(A+3A=1+\frac{1-2}{3}+\frac{-2+3}{3^2}+\frac{3-4}{3^3}+\frac{-4+5}{3^4}+...+\frac{99-100}{3^{99}}-\frac{100}{3^{100}}\)
\(4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-.....+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(4A=(1-\frac{1}{3})+(\frac{1}{3^2}-\frac{1}{3^3})+...+(\frac{1}{3^{98}}-\frac{1}{3^{99}})-\frac{100}{3^{100}}\)
\(4A=\frac{2}{3}+\frac{2}{3^3}+...+\frac{2}{3^{99}}-\frac{100}{3^{100}}\)
\(2A=\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{50}{3^{100}}\)
\(18A=3+\frac{1}{3}+...+\frac{1}{3^{97}}-\frac{450}{3^{100}}\)
\(\Rightarrow 18A-2A=3-\frac{1}{3^{99}}-\frac{450}{3^{100}}+\frac{50}{3^{100}}=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}\)
\(\Leftrightarrow 16A=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}<3\Rightarrow A< \frac{3}{16}\)
Chứng minh rằng:
\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
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
Chứng minh rằng:
\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
Chứng minh rằng:
\(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
\(A=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\\ 3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\\ 3A-A=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\right)\\ 2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\\ 6A-2A=\left(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\\ 4A=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\\ A=\dfrac{3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}}{4}=\dfrac{3}{4}-\dfrac{\dfrac{101}{3^{99}}}{4}-\dfrac{\dfrac{100}{3^{100}}}{4}< \dfrac{3}{4}\)
Vậy ...