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 \(p^{\frac{4}{3}}=\)
- A \(3^{\frac{1}{4}}\)
- B 3
- C 9
- D \(3^{\frac{3}{4}}\)
Answer & Solution
Correct Answer
(B) 3
Step-by-step Solution
Detailed explanation
We have, \(p=3^{\frac{1}{3}} \cdot 3^{\frac{2}{9}} \cdot 3^{\frac{3}{27}}\) \(=3^{\frac{1}{3}\left[1+\frac{2}{3}+\frac{3}{9}+\ldots . \infty\right]}\) \(=3^{\frac{1}{3}\left[\frac{a}{1-r}+\frac{a r}{(1-r)^{2}}\right]}\)…
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