Question
Download Solution PDF\(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{1^2 + 2^2 + 3^2 + .....n^2}{n^3}\) is equal to
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
\({1^2 + 2^2 + 3^2 + .....n^2} = \frac {n\ (n\ +\ 1)(2n\ +\ 1)}{6}\)
Calculation:
Given:
\(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{1^2 + 2^2 + 3^2 + .....n^2}{n^3}\)
= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{n\ (n\ +\ 1)(2n\ +\ 1)}{6\ n^3}\)
= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{n\ (n\ +\ 1)(2n\ +\ 1)}{6\ n^3}\)
= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{(n\ +\ 1)(2n\ +\ 1)}{6\ n^2}\)
= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{n^2(1\ +\ \frac {1}{n})(2\ +\ \frac {1}{n})}{6\ n^2}\)
= \( {\left (1\ +\ \frac {1}{\infty}\right ) \left (2\ +\ \frac {1}{\infty }\right )}\ \frac 1 6\)
= \(\frac 1 3\)
Last updated on May 6, 2025
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