# Lim N → ∞ 1 2 + 2 2 + 3 2 + . . . + N 2 N 3 - Mathematics

MCQ

$\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + . . . + n^2}{n^3}$

• (a) 1

• (b) 1/2

• (c) 1/3

• (d) 0

#### Solution

(c) 1/3

$\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 . . . . . n^2}{n^3}$

$= \lim_{n \to \infty} \frac{\Sigma n^2}{n^3}$

$= \lim_{n \to \infty} \frac{n\left( n + 1 \right) \left( 2n + 1 \right)}{6 n^3}$

$= \lim_{n \to 0} \frac{\left( n + 1 \right) \left( 2n + 1 \right)}{6 n^2}$

$\text{Dividing the numerator and the denominator by n}^2 , we get:$

$\lim_{n \to \infty} \frac{\frac{\left( n + 1 \right)}{n} \times \frac{\left( 2n + 1 \right)}{n}}{6}$

$= \lim_{n \to \infty} \frac{\left( 1 + \frac{1}{n} \right) \left( 2 + \frac{1}{n} \right)}{6}$

$\Rightarrow \frac{2}{6} = \frac{1}{3}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Q 1 | Page 77