Answer:
[tex]\large\boxed{\sum\limits_{n=1}^6400\left(\dfrac{1}{2}\right)^n=393.75}[/tex]
Step-by-step explanation:
The formula of a sum of terms of a geometric sequence:
[tex]S_n=a_1\cdot\dfrac{1-r^n}{1-r}[/tex]
We have:
[tex]\sum\limits^6_{n=1}400\left(\dfrac{1}{2}\right)^n\to a_n=400\left(\dfrac{1}{2}\right)^n[/tex]
Put n = 1:
[tex]a_1=400\left(\dfrac{1}{2}\right)^1=400\left(\dfrac{1}{2}\right)=200\\\\r=\dfrac{a_{n+1}}{a_n}[/tex]
[tex]a_n=400\left(\dfrac{1}{2}\right)^n\\\\a_{n+1}=400\left(\dfrac{1}{2}\right)^{n+1}\\\\r=\dfrac{a_{n+1}}{a_n}=a_{n+1}:a_n=\bigg(400\left(\frac{1}{2}\right)^{n+1}\bigg):\bigg(400\left(\frac{1}{2}\right)^n\bigg)\qquad\text{use}\ a^n:a^m=a^{n-m}\\\\=\left(\dfrac{1}{2}\right)^{n+1-n}=\left(\dfrac{1}{2}\right)^1=\dfrac{1}{2}[/tex]
Substitute:
[tex]a_1=200,\ r=\dfrac{1}{2},\ n=6\\\\S_6=200\cdot\dfrac{1-\left(\frac{1}{2}\right)^6}{1-\frac{1}{2}}=200\cdot\dfrac{1-\frac{1}{64}}{\frac{1}{2}}=200\cdot\dfrac{63}{64}\cdot\dfrac{2}{1}=393.75[/tex]
