$\text { Rewrite } \sum_{i=1}^{n}\left[\frac{12}{n}+\left(\frac{11 i}{n^{2}}\right)\right]\left(\frac{8 i}{n}\right) \text { as a rational function } S(n) \text { and find } \lim _{n \rightarrow \infty} S(n) \text {. }$

A)$S(n)=\frac{4[36 n+11(n+1)]}{3 n^{2}}, \lim _{n \rightarrow \infty} S(n)=\frac{232}{3}$
B)$S(n)=\frac{4[36 n(n+1)+11(n+1)(2 n+1)]}{3 n^{2}}, \lim _{n \rightarrow \infty} S(n)=\frac{232}{3}$
C)$S(n)=\frac{4[36 n(n+1)+11(n+1)(2 n+1)]}{3 n^{2}}, \lim _{n \rightarrow \infty} S(n)=\frac{5}{6}$
D)$S(n)=\frac{4[36+11 n]}{3 n}, \lim _{n \rightarrow \infty} S(n)=\frac{232}{3}$
E)$S(n)=\frac{4[36 n(n+1)+11(n+1)(2 n+1)]}{3 n^{2}}, \text { limit does not exist }$

Question

\text { Rewrite } \sum_{i=1}^{n}\left[\frac{12}{n}+\left(\frac{11 i}{n^{2}}\right) \right]\left(\frac{8 i}{n}\right) \text { as a rational function } S(n) \text { and find } \lim _{n \rightarrow \infty} S(n) \text {. }

A)$S(n)=\frac{4[36 n+11(n+1)]}{3 n^{2}}, \lim _{n \rightarrow \infty} S(n)=\frac{232}{3}$
B)$S(n)=\frac{4[36 n(n+1)+11(n+1)(2 n+1)]}{3 n^{2}}, \lim _{n \rightarrow \infty} S(n)=\frac{232}{3}$
C)$S(n)=\frac{4[36 n(n+1)+11(n+1)(2 n+1)]}{3 n^{2}}, \lim _{n \rightarrow \infty} S(n)=\frac{5}{6}$
D)$S(n)=\frac{4[36+11 n]}{3 n}, \lim _{n \rightarrow \infty} S(n)=\frac{232}{3}$
E)$S(n)=\frac{4[36 n(n+1)+11(n+1)(2 n+1)]}{3 n^{2}}, \text { limit does not exist }$

Quizplus · Accepted Answer

The correct expression for $S(n)$ is obtained by distributing and summing the series, then simplifying. The limit as $n \rightarrow \infty$ of $S(n)$ is $\frac{232}{3}$, which is found by dividing the leading coefficients of the numerator and denominator.