Q71 / 155 · WBJEE 2022 · Mathematics

Let $a_n=\left(1^2+2^2+\ldots n^2\right)^n$ and $b_n=n^n(n !)$. Then

  • A
    $a_n < b_n \forall n$
  • Correct answer
    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

Solution

Using PMI $\quad a_n>b_n \quad\left[{ }^* a_n>b_n \forall n \geq 2\right.$ as $\left.a_1=b_1\right]$

Hence, the correct answer is option: 'B'.

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