Une Somme Difficile

We might think of applying the Generalized Binomial Theorem , but it can't be applied directly.

So, need to do some modifications with the series so that we can easily see some binomial pattern.

$x=1+\frac{1}{12}+\frac{1(4)}{2!}\frac{1}{12^2}+\frac{1(4)(7)}{3!}\frac{1}{12^3}+\frac{1(4)(7)(10)}{4!}\frac{1}{12^4}+\ldots$

We'll break $12^x$ ( $x$ is any arbitrary power of 12) into $4^x \times 3^x$ and take $3^x$ in the numerator. Observe carefully -

$x=1+\frac{1}{3}\times \frac{1}{4}+\frac{\frac{1}{3}\times \frac{4}{3}}{2!}\frac{1}{4^2}+\frac{\frac{1}{3}\times \frac{4}{3}\times \frac{7}{3}}{3!}\frac{1}{4^3}+\ldots$

$x=1+\frac{-1}{3}\times \frac{-1}{4}+\frac{\frac{-1}{3}\times (\frac{-1}{3}-1)}{2!}\frac{1}{4^2}+\frac{\frac{-1}{3}\times (\frac{-1}{3})\times (\frac{-1}{3}-2)}{3!}\frac{-1}{4^3}+\ldots$

Now, this is nothing but expansion of $x = (1-\frac{1}{4})^{\frac{-1}{3}} \Rightarrow (\frac{3}{4})^{\frac{-1}{3}} \Rightarrow (\frac{4}{3})^{\frac{1}{3}}$

Hence, $x = (\frac{4}{3})^{\frac{1}{3}}$ . Therefore, $3x^3 = \boxed{4}$