Rewrite $\sum _ { i = 1 } ^ { n } \left[ \frac { 3 } { n } + \left( \frac { 4 i } { n ^ { 2 } } \right) \right] \left( \frac { 4 i } { n } \right)$ as a rational function S(n)and find $\lim _ { n \rightarrow \infty } S ( n )$ .
A) $S ( n ) = \frac { 2 [ 9 n ( n + 1 ) + 4 ( n + 1 ) ( 2 n + 1 ) ] } { 3 n ^ { 2 } } \text {, limit dos not exist }$
B) $S ( n ) = \frac { 2 [ 9 + 4 n ] } { 3 n } , \lim _ { n \rightarrow \infty } S ( n ) = \frac { 34 } { 3 }$
C) $S ( n ) = \frac { 2 [ 9 n + 4 ( n + 1 ) ] } { 3 n ^ { 2 } } , \lim _ { n \rightarrow \infty } S ( n ) = \frac { 34 } { 3 }$
D) $S ( n ) = \frac { 2 [ 9 n ( n + 1 ) + 4 ( n + 1 ) ( 2 n + 1 ) ] } { 3 n ^ { 2 } } , \lim _ { n \rightarrow \infty } S ( n ) = \frac { 5 } { 6 }$
E) $S ( n ) = \frac { 2 [ 9 n ( n + 1 ) + 4 ( n + 1 ) ( 2 n + 1 ) ] } { 3 n ^ { 2 } } , \lim _ { n \rightarrow \infty } S ( n ) = \frac { 34 } { 3 }$

Question

Quizplus · Accepted Answer

The given sum can be rewritten and simplified using the formula for the sum of the first n natural numbers and the sum of the squares of the first n natural numbers. The correct rational function S(n) is given in option E, and the limit as n approaches infinity is $\frac{34}{3}$, which is found by simplifying the expression and dividing the highest degree terms in the numerator and denominator.

Rewrite ∑i=1n[3n+(4in2)](4in)\sum _ { i = 1 } ^ { n } \left[ \frac { 3 } { n } + \left( \frac { 4 i } { n ^ { 2 } } \right) \right] \left( \frac { 4 i } { n } \right)∑i=1n​[n3​+(n24i​)](n4i​)

Rewrite $\sum _ { i = 1 } ^ { n } \left[ \frac { 3 } { n } + \left( \frac { 4 i } { n ^ { 2 } } \right) \right] \left( \frac { 4 i } { n } \right)$