AP EAMCET · Maths · Sequences and Series
For all \(n \in N, \frac{3^n-1}{2} \geq\)
- A \(n^2\left(2^{n / 2}\right)\)
- B \(n^2\left(3^{\frac{n-1}{2}}\right)\)
- C \(n^3\left(3^{\frac{n-1}{2}}\right)\)
- D \(n\left(3^{\frac{n-1}{2}}\right)\)
Answer & Solution
Correct Answer
(D) \(n\left(3^{\frac{n-1}{2}}\right)\)
Step-by-step Solution
Detailed explanation
For \(n=1\): \(\frac{3^1-1}{2} = 1\) Option 1: \(1^2(2^{1/2}) = \sqrt{2} \approx 1.41\). \(1 \not\geq 1.41\) Option 2: \(1^2(3^0) = 1\). \(1 \geq 1\) Option 3: \(1^3(3^0) = 1\). \(1 \geq 1\) Option 4: \(1(3^0) = 1\). \(1 \geq 1\) For \(n=2\): \(\frac{3^2-1}{2} = 4\) Option 2:…
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