Answer:
2
Step-by-step explanation:
The formula of the sum of a geometric sequence:
[tex]S_n=\dfrac{a_1(1-r^n)}{1-r}[/tex]
We have
[tex]\sum\limits_{x=0}^{15}2\left(\dfrac{1}{2}\right)^x\to a_n=2\left(\dfrac{1}{2}\right)^n[/tex]
Calculate the common ratio:
[tex]r=\dfrac{a_{n+1}}{a_n}[/tex]
[tex]a_{n+1}=2\left(\dfrac{1}{2}\right)^{n+1}[/tex]
Substitute:
[tex]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}[/tex]
Calculate the sum.
[tex]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[/tex]
