Question

Find the sum of the first 9 terms of the following series, to the nearest integer. 3, 15/4, 75/16...

Answer 1

The series is a geometric series which is governed by the following formula:

`\begin{gathered} T_n=ar^(n-1) \\ \text{where:} \\ T_n=nth\text{ term of the series} \\ a=\text{ first term of the series} \\ r=\text{common ratio of the series} \end{gathered}`

Thus, we have that:

a = 3 ,

To find r, we have to solve for it as follows:

`\begin{gathered} \text{since }T_n=ar^(n-1) \\ \text{Thus, the second term, }T_2\text{ is:} \\ T_2=ar^(2-1) \\ \Rightarrow T_2=ar^1 \\ \sin ce\text{ }T_2=(15)/(4),\text{ and a =3. we have:} \\ \Rightarrow T_2=ar \\ (15)/(4)=3* r \\ 3* r=(15)/(4) \\ \Rightarrow r=((15)/(4))/(3)=(15)/(4)*(1)/(3)=(5)/(4) \\ \Rightarrow r=(5)/(4) \end{gathered}`

Thus, r = 5/4

- Now the sum of the first n terms of a geometric series is given by the formula:

`S_n=(a(r^n-1))/(r-1)`

Thus, the sum of the first 9 terms is:

`S_9=(3*((5)/(4)-1))/((5)/(4)-1)`

simplifying the above gives:

`\begin{gathered} S_9=(3*(((5)/(4))^9-1))/((5)/(4)-1) \\ \Rightarrow S_9=(3*((1953125)/(262144)-1))/((5)/(4)-(4)/(4))=(3*((1953125-262144)/(262144)))/((5-4)/(4))=(3*(1690981)/(262144))/((1)/(4)) \\ \Rightarrow S_9=3*4*(1690981)/(262144)=(20291772)/(262144)=77.4 \\ \Rightarrow S_9=77\text{ (to the nearest integer)} \end{gathered}`

Therefore, the sum of the first 9 terms of the series is 77 (to the nearest integer)