## Calculus: Early Transcendentals 8th Edition

$\frac{195}{4}$
Find the limit $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{3}{n}[(1+\frac{3i}{n})^{3}-2(1+\frac{3i}{n})]$ $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{3}{n}[(1+\frac{3i}{n})^{3}-2(1+\frac{3i}{n})]=\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{3}{n}[(-1+\frac{3i}{n}+\frac{27i^{2}}{n^{2}}+\frac{27i^{3}}{n^{3}}]$ $\lim\limits_{n \to \infty}[-\frac{3}{n}\sum\limits_{i =1}^{n}1+\frac{9}{n^{2}}\sum\limits_{i =1}^{n}i+\frac{81}{n^{3}}\sum\limits_{i =1}^{n}i^{2}+\frac{81}{n^{4}}\sum\limits_{i =1}^{n}i^{3}$ $\sum \limits_{i =1}^{n}i^{2}=[\frac{n(n+1)(2n+1)}{6}]$,$\sum \limits_{i =1}^{n}i^{3}=[\frac{n(n+1)}{2}]^{2}$ Thus, $=\lim\limits_{n \to \infty}[-3+\frac{9}{2}+\frac{9}{2n}+27+\frac{81}{2n}+\frac{27}{2n^{2}}+\frac{81}{4}+\frac{81}{2n}+\frac{81}{4n^{2}}]$ $=\lim\limits_{n \to \infty}[-3+\frac{9}{2}+0+27+0+0+\frac{81}{4}+0+0]$ Hence, $\lim\limits_{n \to \infty}\sum\limits_{i =1}^{n}\frac{3}{n}[(1+\frac{3i}{n})^{3}-2(1+\frac{3i}{n})]=\frac{195}{4}$