Answer 1

(c) 1/3 

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

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

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

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

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

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

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

\[ \Rightarrow \frac{2}{6} = \frac{1}{3}\]