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![](https://rs.olm.vn/images/avt/0.png?1311)
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:...........
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt :
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+........+\frac{100}{3^{100}}\)
\(\Leftrightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+.....+\frac{100}{3^{99}}\)
\(\Leftrightarrow3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+....+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+....+\frac{100}{3^{100}}\right)\)
\(\Leftrightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
Đặt : \(H=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\) \(\Leftrightarrow2A=H-\frac{100}{3^{100}}\)
\(\Leftrightarrow3H=3+1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)
\(\Leftrightarrow3H-H=\left(4+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)\)
\(\Leftrightarrow2H=3-\frac{1}{3^{99}}\)
\(\Leftrightarrow H=\frac{3-\frac{1}{99}}{2}\)
\(\Leftrightarrow2A=\frac{3-\frac{1}{3^{99}}}{2}-\frac{100}{3^{100}}\)
\(\Leftrightarrow A=\frac{1-\frac{1}{3^{99}}}{2}-\frac{100}{2.3^{100}}\)
\(\Leftrightarrow A< \frac{3}{4}\left(đpcm\right)\)
Mà sao bạn tức giận thế nhỉ, mọi khi có thể đâu. Khổ thật.
Nguyễn Văn Đạt
![](https://rs.olm.vn/images/avt/0.png?1311)
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!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)