Find the first and third partial sums, s1 and s3, for the sequence
-$a _ { n } = \frac { n ^ { 2 } } { n + 4 }$
A) $\frac { 1 } { 5 } , \frac { 9 } { 7 }$
B) $\frac { 1 } { 5 } , \frac { 1 } { 15 }$
C) $\frac { 1 } { 5 } , \frac { 226 } { 105 }$
D) $\frac { 1 } { 5 } , \frac { 7 } { 5 }$

Question

Quizplus · Accepted Answer

To find the first and third partial sums, $s_1$ and $s_3$, for the sequence $a_n = \frac{n^2}{n + 4}$, we calculate the values of the sequence for $n = 1, 2, 3$ and then sum them up accordingly.For $n = 1$, $a_1 = \frac{1^2}{1 + 4} = \frac{1}{5}$, so $s_1 = \frac{1}{5}$.For $n = 2$, $a_2 = \frac{2^2}{2 + 4} = \frac{4}{6} = \frac{2}{3}$.For $n = 3$, $a_3 = \frac{3^2}{3 + 4} = \frac{9}{7}$.Thus, $s_3 = a_1 + a_2 + a_3 = \frac{1}{5} + \frac{2}{3} + \frac{9}{7}$.To find $s_3$, we need to find a common denominator and sum these fractions: The common denominator for 5, 3, and 7 is 105, so we convert each fraction:$\frac{1}{5} = \frac{21}{105}$, $\frac{2}{3} = \frac{70}{105}$, $\frac{9}{7} = \frac{135}{105}$.Adding these together, $s_3 = \frac{21 + 70 + 135}{105} = \frac{226}{105}$.Therefore, the correct answer is $s_1 = \frac{1}{5}$ and $s_3 = \frac{226}{105}$, which matches choice C.

Find the First and Third Partial Sums, S1 and S3 an=n2n+4a _ { n } = \frac { n ^ { 2 } } { n + 4 }an​=n+4n2​

Find the First and Third Partial Sums, S1 and S3 $a _ { n } = \frac { n ^ { 2 } } { n + 4 }$