If sum of the first 21 terms of the series $$\log_{9^{1/2}} x + \log_{9^{1/3}} x + \log_{9^{1/4}} x + \ldots$$ where $$x > 0$$ is 504, then $$x$$ is equal to:

The general term of the series is $$\log_{9^{1/(k+1)}} x$$ for $$k = 1, 2, \ldots, 21$$.

Using the change of base formula, $$\log_{9^{1/(k+1)}} x = \frac{\log x}{\log 9^{1/(k+1)}} = \frac{\log x}{\frac{1}{k+1}\log 9} = (k+1)\log_9 x$$.

So the $$k$$-th term (for $$k = 1$$ to $$21$$) is $$(k+1)\log_9 x$$.

The sum of the first 21 terms is: $$\log_9 x \cdot \sum_{k=1}^{21}(k+1) = \log_9 x \cdot (2 + 3 + 4 + \cdots + 22).$$

$$\sum_{k=2}^{22} k = \frac{22 \cdot 23}{2} - 1 = 253 - 1 = 252.$$

So $$252 \log_9 x = 504$$, giving $$\log_9 x = 2$$, hence $$x = 9^2 = 81$$.