Question

What is the sum of the geometric series 3∑i=1(8(1/4)^i-1)?
A.) 21
B.) 21/2
C.) 10
D.) 21/8

Answer 1

The answer should be b

Answer 2

Answer:

$\frac{21}{2}$

Step-by-step explanation:

Geometric series is a sequence of form $a_1,a_2,a_3,...,a_n$, such that $\frac{a_{n-1}}{a_{n}}$ is constant .

To find : sum of the geometric series  $\sum_{i=1}^{3}\left ( 8\left ( \frac{1}{4} \right )^{i-1} \right )$

Solution:

For i = 1:

Put i = 1 in the term $8\left ( \frac{1}{4} \right )^{i-1}$

$\left ( 8\left ( \frac{1}{4} \right )^{i-1} \right )=\left ( 8\left ( \frac{1}{4} \right )^{1-1} \right )=8(1)=8$

( here, formula used: $a^0=1$ )

For i = 2:

Put i= 2 in the term $8\left ( \frac{1}{4} \right )^{i-1}$

$\left ( 8\left ( \frac{1}{4} \right )^{i-1} \right )=\left ( 8\left ( \frac{1}{4} \right )^{2-1} \right )=\frac{8}{4}=2$

For i = 3:

Put i = 3 in the term $8\left ( \frac{1}{4} \right )^{i-1}$

$\left ( 8\left ( \frac{1}{4} \right )^{i-1} \right )=\left ( 8\left ( \frac{1}{4} \right )^{3-1} \right )=\frac{8}{16}=\frac{1}{2}$

Therefore,

$\sum_{i=1}^{3}\left ( 8\left ( \frac{1}{4} \right )^{i-1} \right )=\left ( 8\left ( \frac{1}{4} \right )^{1-1} \right )+\left ( 8\left ( \frac{1}{4} \right )^{2-1} \right )+\left ( 8\left ( \frac{1}{4} \right )^{3-1} \right )\\=8+2+\frac{1}{2}\\=\frac{16+4+1}{2}\\=\frac{21}{2}$

So, option 2. is correct