WBJEE · Maths · Sequences and Series
Let \(a_n=\left(1^2+2^2+\ldots n^2\right)^n\) and \(b_n=n^n(n !)\). Then
- A \(a_n < b_n \forall n\)
- B \(a_n>b_n \forall n\)
- C \(a_n=b_n\) for infinitely many \(n\)
- D \(a_n < b_n\) if \(n\) be even and \(a_n>b_n\) if \(n\) be odd
Answer & Solution
Correct Answer
(B) \(a_n>b_n \forall n\)
Step-by-step Solution
Detailed explanation
Using PMI \(\quad a_n>b_n \quad\left[{ }^* a_n>b_n \forall n \geq 2\right.\) as \(\left.a_1=b_1\right]\)
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