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Mình ngu lắm dân trần đăng ninh chuyên anh mà làm sao giỏi toán được
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-.....+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow 4A=A+3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(12A=3-1+\frac{1}{3}-\frac{1}{3^2}+...-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
$\Rightarrow 4A+12A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}<3$
$\Rightarrow 16A< 3$
$\Rightarrow A< \frac{3}{16}$
Chứng minh rằng:
a) 1/2-1/4+1/8-1/16+1/32-1/64<1/3
b) 1/3-2/3^2+3/3^3-3/3^4+...+99/3^99-100/3^100<3/16
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A=1/3 - 2/3^2+3/3^3 - 4/3^4+ ... - 100/3^100
=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99
=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100
=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99
=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...
<=>16A=3-101/3^99-100/3^100
<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16
Suy ra A<3/16