Đặt `A = \frac{1}{3^2} + \frac{3}{3^4} + \frac{5}{3^6} + \cdots + \frac{99}{3^{100}}`

`\frac{1}{9}A = \frac{1}{3^4} + \frac{3}{3^6} + \frac{5}{3^8} + \cdots + \frac{97}{3^{100}} + \frac{99}{3^{102}}`

`A - \frac{1}{9}A = (\frac{1}{3^2} + \frac{3}{3^4} + \frac{5}{3^6} + \cdots + \frac{99}{3^{100}}) - (\frac{1}{3^4} + \frac{3}{3^6} + \frac{5}{3^8} + \cdots + \frac{97}{3^{100}} + \frac{99}{3^{102}})`

`\frac{8}{9}A = \frac{1}{3^2} + (\frac{3}{3^4} - \frac{1}{3^4}) + (\frac{5}{3^6} - \frac{3}{3^6}) + \cdots + (\frac{99}{3^{100}} - \frac{97}{3^{100}}) - \frac{99}{3^{102}}`

`\frac{8}{9}A = \frac{1}{3^2} + \frac{2}{3^4} + \frac{2}{3^6} + \cdots + \frac{2}{3^{100}} - \frac{99}{3^{102}}`

`\frac{8}{9}A = \frac{1}{9} + 2(\frac{1}{3^4} + \frac{1}{3^6} + \cdots + \frac{1}{3^{100}}) - \frac{99}{3^{102}}`

Đặt `B = \frac{1}{3^4} + \frac{1}{3^6} + \cdots + \frac{1}{3^{100}}`

`B = \frac{1}{3^4} \cdot \frac{1 - (\frac{1}{3^2})^{49}}{1 - \frac{1}{3^2}}`

`= \frac{1}{81} \cdot \frac{1 - (\frac{1}{9})^{49}}{\frac{8}{9}}`

`= \frac{1}{81} \cdot \frac{9}{8} \cdot (1 - (\frac{1}{9})^{49}) `

`= \frac{1}{72} \cdot (1 - (\frac{1}{9})^{49})`

`=> \frac{8}{9}A = \frac{1}{9} + 2 \cdot \frac{1}{72} \cdot (1 - (\frac{1}{9})^{49}) - \frac{99}{3^{102}}`

`\frac{8}{9}A = \frac{1}{9} + \frac{1}{36} \cdot (1 - (\frac{1}{9})^{49}) - \frac{99}{3^{102}}`

`\frac{8}{9}A = \frac{1}{9} + \frac{1}{36} - \frac{1}{36} \cdot (\frac{1}{9})^{49} - \frac{99}{3^{102}}`

`\frac{8}{9}A = \frac{5}{36} - (\frac{1}{36} \cdot (\frac{1}{9})^{49} + \frac{99}{3^{102}})`

Vì `(\frac{1}{36} \cdot (\frac{1}{9})^{49} + \frac{99}{3^{102}}) > 0`

`\frac{8}{9}A < \frac{5}{36}`

`A < \frac{5}{36} \cdot \frac{9}{8}`

`A < \frac{5 \cdot 9}{36 \cdot 8}`

`A < \frac{5}{32}`

Vậy `A < \frac{5}{32}`