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COMEDK · Maths · 5. Sequences and Series

If \(p=3^{\frac{1}{3}} \cdot 3^{\frac{2}{9}} \cdot 3^{\frac{3}{27}} \ldots \infty\), then \(p^{\frac{4}{3}}=\)

  1. A \(3^{\frac{1}{4}}\)
  2. B 3
  3. C 9
  4. D \(3^{\frac{3}{4}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) 3

Step-by-step Solution

Detailed explanation

We have, \(p=3^{\frac{1}{3}} \cdot 3^{\frac{2}{9}} \cdot 3^{\frac{3}{27}}\) \(=3^{\frac{1}{3}\left[1+\frac{2}{3}+\frac{3}{9}+\ldots . \infty\right]}\) \(=3^{\frac{1}{3}\left[\frac{a}{1-r}+\frac{a r}{(1-r)^{2}}\right]}\)…