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Phần C đề thiếu
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)
\(\Rightarrow4D=3-\frac{203}{3^{100}}\)
\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)
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!!!
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)
\(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3C=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^2}+...+\frac{99}{3^{89}}-\frac{100}{3^{99}}\)
\(\Rightarrow4C=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4C< 1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\left(1\right)\)
Đặt: \(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(\Rightarrow3B=2+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(4B=B+3B=3-\frac{1}{3^{99}}< 3\)
\(\Rightarrow B< \frac{3}{4}\left(2\right)\)
Từ: \(\left(1\right)\left(2\right)\Rightarrow4C< B< \frac{3}{4}\)
\(\Rightarrow C< \frac{3}{16}\left(đpcm\right)\)
(Đánh nhanh quá sai chỗ nào thông cảm nha :))
\(M=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}.\)
\(3M=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+....+\frac{100}{3^{99}}\)
\(3M-M=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}-\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}\)
\(2M=1+\left(\frac{2}{3}-\frac{1}{3}\right)+\left(\frac{3}{3^2}-\frac{2}{3^2}\right)+...+\left(\frac{100}{3^{99}}-\frac{100}{3^{99}}\right)-\frac{100}{3^{100}}\)
\(2M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
Đặt :
\(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\)
\(3N=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{98}}\)
\(3N-N=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(2N=1-\frac{1}{3^{99}}\)
\(N=\frac{1-\frac{1}{3^{99}}}{2}\)
Thay vào ta có :
\(2M=1+\frac{1-\frac{1}{3^{99}}}{2}-\frac{100}{3^{100}}\)
\(2M=1+\frac{1}{2}-\frac{1}{2\times3^{100}}-\frac{100}{3^{100}}< 1+\frac{1}{2}=\frac{3}{2}\)
Ta có : \(\frac{3}{4}< \frac{3}{2}\)
\(\Rightarrow\)\(2M=1+\frac{1}{2}-\frac{1}{2\times3^{100}}-\frac{100}{3^{100}}< \frac{3}{4}\)
\(\Rightarrow\)\(M< \frac{3}{4}\)
* Sai thì xin lỗi ạ ! *