0,(09)+0,(18)+0,(27)+…+0,(81)+0,(90)

To evaluate the sum \(0.(09) + 0.(18) + 0.(27) + \ldots + 0.(90)\), first, we need to express the repeating decimal in fractional form.

1. Convert each repeating decimal:

   • \(0.(09) = \frac{9}{99} = \frac{1}{11}\)
   • \(0.(18) = \frac{18}{99} = \frac{2}{11}\)
   • \(0.(27) = \frac{27}{99} = \frac{3}{11}\)
   • \(0.(36) = \frac{36}{99} = \frac{4}{11}\)
   • \(0.(45) = \frac{45}{99} = \frac{5}{11}\)
   • \(0.(54) = \frac{54}{99} = \frac{6}{11}\)
   • \(0.(63) = \frac{63}{99} = \frac{7}{11}\)
   • \(0.(72) = \frac{72}{99} = \frac{8}{11}\)
   • \(0.(81) = \frac{81}{99} = \frac{9}{11}\)
   • \(0.(90) = \frac{90}{99} = \frac{10}{11}\)

2. The series can now be rewritten: \[ \frac{1}{11} + \frac{2}{11} + \frac{3}{11} + \frac{4}{11} + \frac{5}{11} + \frac{6}{11} + \frac{7}{11} + \frac{8}{11} + \frac{9}{11} + \frac{10}{11} \]

3. Factor out \(\frac{1}{11}\): \[ \frac{1}{11} \left(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10\right) \]

4. Calculate the sum inside the parentheses: \[ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = \frac{10 \cdot (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55 \]

5. Now substitute back into the formula: \[ \frac{1}{11} \cdot 55 = \frac{55}{11} = 5 \]

So the final result of the sum \(0.(09) + 0.(18) + 0.(27) + \ldots + 0.(90)\) is \(\boxed{5}\).