Rewrite $\sum _ { i = 1 } ^ { n } \left[ \frac { 12 } { n } + \left( \frac { 11 i } { n ^ { 2 } } \right) \right] \left( \frac { 8 i } { n } \right)$ as a rational function $S ( n )$ and find $\lim _ { n \rightarrow \infty } S ( n )$.
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) $S ( n ) = \frac { 4 [ 36 n ( n + 1 ) + 11 ( n + 1 ) ( 2 n + 1 ) ] } { 3 n ^ { 2 } }$, limit does not exist

Question

Quizplus · Accepted Answer

First, distribute the terms inside the sum and simplify. The sum of $i$ from 1 to $n$ is $\frac{n(n+1)}{2}$, and the sum of $i^2$ from 1 to $n$ is $\frac{n(n+1)(2n+1)}{6}$. Substituting these into the given expression and simplifying yields $S(n)$ as in choice B. The limit as $n$ approaches infinity of $S(n)$ simplifies to $\frac{232}{3}$ after dividing the numerator and denominator by $n^3$ and taking the limit.

Rewrite ∑i=1n[12n+(11in2)](8in)\sum _ { i = 1 } ^ { n } \left[ \frac { 12 } { n } + \left( \frac { 11 i } { n ^ { 2 } } \right) \right] \left( \frac { 8 i } { n } \right)∑i=1n​[n12​+(n211i​)](n8i​)

Rewrite $\sum _ { i = 1 } ^ { n } \left[ \frac { 12 } { n } + \left( \frac { 11 i } { n ^ { 2 } } \right) \right] \left( \frac { 8 i } { n } \right)$