The value of \(\rm 64^{\tfrac{1}{3}}\cdot 64^{\tfrac{1}{9}}\cdot 64^{\tfrac{1}{27}}\ ...\ \infty\) is:

Concept:

Geometric Progression (GP):

• The series of numbers where the ratio of any two consecutive terms is the same is called a Geometric Progression.
• A Geometric Progression of n terms with first term a and common ratio r is represented as:

  a, ar, ar2, ar3, ..., arn-2, arn-1.

• The sum of the first n terms of a GP is: Sn = \(\rm a\left(\dfrac{r^n-1}{r-1}\right)\).
• The sum  ∞ of a GP, when |r| < 1, is: S∞ = \(\rm \dfrac{a}{1-r}\).

Calculation:

Given: \(\rm 64^{\tfrac{1}{3}}\cdot 64^{\tfrac{1}{9}}\cdot 64^{\tfrac{1}{27}}\ ...\ \infty\)

\(\rm \rm 64^{(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ ... \infty)}\)

Let us consider the infinite series  \(\rm \dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ ... \infty\).

Here, a = \(\rm \dfrac{1}{3}\) and r = \(\rm \dfrac{\tfrac{1}{9}}{\tfrac{1}{3}}=\dfrac{1}{3}\).

∴ S∞ = \(\rm \dfrac{a}{1-r}=\dfrac{\tfrac{1}{3}}{1-\tfrac{1}{3}}=\dfrac{\tfrac{1}{3}}{\tfrac{2}{3}}=\dfrac{1}{2}\).

Now, let P = \(\rm \rm 64^{(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ ... \infty)}\)

∴ P = \(\rm 64^{\tfrac{1}{2}}=\sqrt64=8\).