Question

What is the sum of the geometric series rounded to the whole number

Answer 1

Answer:

2

Step-by-step explanation:

The formula of the sum of a geometric sequence:

$S_n=\dfrac{a_1(1-r^n)}{1-r}$

We have

$\sum\limits_{x=0}^{15}2\left(\dfrac{1}{2}\right)^x\to a_n=2\left(\dfrac{1}{2}\right)^n$

Calculate the common ratio:

$r=\dfrac{a_{n+1}}{a_n}$

$a_{n+1}=2\left(\dfrac{1}{2}\right)^{n+1}$

Substitute:

$r=\dfrac{2\left(\frac{1}{2}\right)^{n+1}}{2\left(\frac{1}{2}\right)^n}=\left(\dfrac{1}{2}\right)^{n+1}:\left(\dfrac{1}{2}\right)^n\\\\\text{use}\ a^n:a^m=a^{n-m}\\\\r=\left(\dfrac{1}{2}\right)^{n+1-n}=\left(\dfrac{1}{2}\right)^1=\dfrac{1}{2}$

Calculate the sum.

$a_1=2\left(\dfrac{1}{2}\right)^1=2\left(\dfrac{1}{2}\right)=1;\ r=\dfrac{1}{2},\ n=15\\\\S_{15}=\dfrac{1\left(1-\left(\frac{1}{2}\right)^{15}\right)}{1-\frac{1}{2}}=\dfrac{1-\frac{1}{2^{15}}}{\frac{1}{2}}=\left(1-\dfrac{1}{2^{15}}\right)\left(\dfrac{2}{1}\right)=2-\dfrac{2}{2^{15}}=2-\dfrac{1}{2^{14}}\\\\=2-\dfrac{1}{16384}=1\dfrac{16383}{16384}\approx2$