Question

Find the sum of the geometric sequence. 1, 1/4, 1/16, 1/64, 1/256

A) 341

B) 1/192

C) 1/768

D) 341/256

Answer 1

$1,\ \dfrac{1}{4};\ \dfrac{1}{16};\ \dfrac{1}{64};\ \dfrac{1}{256}\ \text{is geometric sequence}\\\\\text{the ratio is:}\ r=\dfrac{\frac{1}{4}}{1}=\dfrac{1}{4}\\\\\text{The sum of the numbers in a geometric sequence is:}\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\a_1=1;\ r=\dfrac{1}{4}\\\\\text{substitute}\\\\S_5=\dfrac{1\cdot\left[1-\left(\dfrac{1}{4}\right)^5\right]}{1-\dfrac{1}{4}}=\dfrac{1-\dfrac{1}{1024}}{\dfrac{3}{4}}=\dfrac{1023}{1024}\cdot\dfrac{4}{3}=\dfrac{341}{256}\to D)$

Other method:
$1=\dfrac{256}{256}\\\\\dfrac{1}{4}=\dfrac{1\cdot64}{4\cdot64}=\dfrac{64}{256}\\\\\dfrac{1}{16}=\dfrac{1\cdot16}{16\cdot16}=\dfrac{16}{256}\\\\\dfrac{1}{64}+\dfrac{1\cdot4}{64\cdot4}=\dfrac{4}{256}\\\\1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}=\dfrac{256}{256}+\dfrac{64}{256}+\dfrac{16}{256}+\dfrac{4}{256}+\dfrac{1}{256}=\dfrac{341}{256}\to D)$