The value of _ 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 )

Correct answer: A. x 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 = xΣn^ 2 - n Σn^ 2 = x - (6)/( (n + 1)(2n + 1)) (∵ x - 1 ≤ [x] Thus, we have, x - (6)/((n + 1)(2n + 1)) Now, we have, _ n → ∞ x - (6)/((n + 1)(2n + 1)) = x & _ n → ∞ x = x Hence, by Sandwich Theorem, we have _ n → ∞ f(x) = x

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Limits

NTA Abhyas 2020 · Mathematics · Limits

The value of _ n → ∞ [x] + [ 2 ^ 2 x] + [ 3 ^ 2 x] +..... + [n^ 2...

The value of

\lim_{n \to \infty}\dfrac{\left[x\right] + \left[{2}^{2}x\right] + \left[{3}^{2}x\right] +..... + \left[n^{2}x\right]}{{1}^{2} + {2}^{2} + {3}^{2} +.... + n^{2}}

is equal to (where \left[x\right]

represents the greatest integer part of

x$)

Options

A $x$
B $2x$
C $\dfrac{x}{2}$
D $\dfrac{x}{6}$

Correct answer

A. $x$

Step-by-step solution

Let

f\left(x\right) = \dfrac{\left[x\right] + \left[{2}^{2}x\right] + \left[{3}^{2}x\right] +... + \left[n^{2}x\right]}{{1}^{2} + {2}^{2} + {3}^{2} +... + n^{2}}

Now, we have,

f\left(x\right) \le \dfrac{x + {2}^{2}x + {3}^{2}x +... + n^{2}x}{{1}^{2} + {2}^{2} + {3}^{2} +... + n^{2}} = x

and,

f\left(x\right) > \dfrac{\left(x - 1\right) + \left({2}^{2}x - 1\right) + \left({3}^{2}x - 1\right) +... + \left(n^{2}x - 1\right)}{{1}^{2} + {2}^{2} + {3}^{2} +... + n^{2}}

= \dfrac{xΣn^{2} - n}{Σn^{2}} = x - \dfrac{6}{ \left(n + 1\right)\left(2n + 1\right)} \left(∵ x - 1 \le \left[x\right] < x, \forall x \in R\right)

Thus, we have,

x - \dfrac{6}{\left(n + 1\right)\left(2n + 1\right)} < f\left(x\right) \le x

Now, we have,

\lim_{n \to \infty}x - \dfrac{6}{\left(n + 1\right)\left(2n + 1\right)} = x & \lim_{n \to \infty}x = x

Hence, by Sandwich Theorem, we have

\lim_{n \to \infty}f\left(x\right) = x

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