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M = 1/3 + 2/3² + 3/3³ + 4/3^4 + ... + 100/3^100
=> 3M= 1 + 2/3 + 3/3² + 4/3³ + .... + 100/3^99
=> 3M-M = 1 + ﴾2/3 ‐ 1/3﴿ + ﴾3/3² ‐ 2/3²﴿ +...+ ﴾100/3^99 ‐ 99/3^99﴿ ‐ 100/3^100
=> 2M= 1+ 1/3 + 1/3² + 1/3³ +...+ 1/3^99 ‐ 100/3^100
Đặt N = 1/3 + 1/3² + 1/3³ +...+ 1/3^99
=> 3N = 1 + 1/3 + 1/3² + 1/3³ +...+ 1/3^98
=> 2N = 1 ‐ 1/3^99
=> N = ﴾1 ‐ 1/3^99﴿/2
Thay vào 2M
=> 2M= 1+ 1/2 ‐ 1/﴾2x3^99﴿ ‐ 100/3^100 < 1+ 1/2 = 3/2
=> M < 3/4
vậy...
Bài này công nhận là dễ , nhưng khi nãy bận ăn cơm , xin lỗi ha!! Hứa lần sau sẽ giải cho!!!
Phần C đề thiếu
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)
\(\Rightarrow4D=3-\frac{203}{3^{100}}\)
\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)
Giúp mình nha. Bài cuối cùng của đề toán dài 36 bài của mình đó
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
Nên từ đây => \(A< 1\) (ĐPCM)
Đặt:
\(4C=3-\frac{203}{3^{100}}.\)
Mà \(3-\frac{203}{3^{100}}< 3\)
\(\Rightarrow4C< 3\)
\(\Rightarrow C< \frac{3}{4}\left(đpcm\right).\)
Vậy \(C< \frac{3}{4}.\)
Chúc bạn học tốt!
hổng khó, marivan2016(mk bít nick thiệt nhưng hổng nói) làm ơn k giùm mk nha cảm ơn nhìu!!!
\(3C=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+....+\frac{100}{3^{99}}.\)
\(2C=3C-C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}-\frac{100}{3^{100}}.\)
\(2C=1+A-\frac{100}{3^{100}}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}=\frac{1}{2}\left(1-\frac{1}{3^{99}}\right)< \frac{1}{2}\)
=>\(2C=1+A-\frac{100}{3^{100}}< 1+\frac{1}{2}=\frac{3}{2}\)
\(C< \frac{3}{4}.\)
CMR:
\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+.....+\frac{100}{3^{100}}< \frac{3}{4}\)
đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)
\(\Rightarrow A=\frac{3-\frac{203}{3^{100}}}{4}\)
\(\Rightarrow A=\frac{3}{4}-\frac{203}{\frac{3^{100}}{4}}\le\frac{3}{4}\left(ĐPCM\right)\)
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)
\(\Rightarrow D=\frac{3-\frac{203}{3^{100}}}{4}=\frac{3}{4}-\frac{203}{3^{100}.4}< \frac{3}{4}\left(đpcm\right)\)
Vậy \(D< \frac{3}{4}\)
Nguồn: @Dekisugi Hidetoshi
A = \(\frac{1}{3}\) + \(\frac{2}{3^2}\) + \(\frac{3}{3^3}\) + \(\frac{4}{3^4}\) +....+ \(\frac{100}{3^{100}}\)
3A = 1 + \(\frac{2}{3}\) + \(\frac{3}{3^2}\) + \(\frac{4}{3^3}\) +...+ \(\frac{100}{3^{99}}\)
\(\Rightarrow\) 3A - A = 1+ \(\left(\frac{2}{3}-\frac{1}{3}\right)\) + \(\left(\frac{3}{3^2}-\frac{2}{3^2}\right)\) + ... + \(\left(\frac{100}{3^{99}}-\frac{99}{3^{99}}\right)\) - \(\frac{100}{3^{100}}\)
2A =1+ \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
Đặt B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{99}}\)
\(\Rightarrow\) 3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow\) 2B = \(1-\frac{1}{3^{99}}\)
\(\Rightarrow\) \(B=\left(1-\frac{1}{3^{99}}\right):2\)
Thay 2A = 1 + \(\frac{1}{2}\) - \(\left(1-\frac{2}{3^{99}}\right)\) - \(\frac{100}{3^{100}}\) < 1 + \(\frac{1}{2}\) = \(\frac{3}{2}\)
Vậy A < \(\frac{3}{4}\)
Vậy:...........
2M=... như trên là sao hả bạn