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NTA Abhyas JEE Main2020MathematicsLimitsPractice

The value of l i m n → ∞ x + 2 2 x + 3 2 x + . . . . . + n 2 x 1 2 + 2 2 + 3 2 + . . . . + n 2 is equal to (where x represents the greatest integer part of x )

Options

  1. Ax
  2. B2 x
  3. Cx 2
  4. Dx 6

Correct answer

A. x

Step-by-step solution

Let f x = x + 2 2 x + 3 2 x + . . . + n 2 x 1 2 + 2 2 + 3 2 + . . . + n 2 Now, we have, f x ≤ x + 2 2 x + 3 2 x + . . . + n 2 x 1 2 + 2 2 + 3 2 + . . . + n 2 = x and, f x > x - 1 + 2 2 x - 1 + 3 2 x - 1 + . . . + n 2 x - 1 1 2 + 2 2 + 3 2 + . . . + n 2 = x Σ n 2 - n Σ n 2 = x - 6 n + 1 2 n + 1 ∵ x - 1 ≤ x < x ,   ∀ x ∈ R Thus, we have, x - 6 n + 1 2 n + 1 < f x ≤ x Now, we have, l i m n → ∞ x - 6 n + 1 2 n + 1 = x     &  

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