Question

Find the sum of the geometric. Remember to show work and explain.​

Answer 1

Answer:

$\large\boxed{\sum\limits_{n=1}^6400\left(\dfrac{1}{2}\right)^n=393.75}$

Step-by-step explanation:

The formula of a sum of terms of a geometric sequence:

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

We have:

$\sum\limits^6_{n=1}400\left(\dfrac{1}{2}\right)^n\to a_n=400\left(\dfrac{1}{2}\right)^n$

Put n = 1:

$a_1=400\left(\dfrac{1}{2}\right)^1=400\left(\dfrac{1}{2}\right)=200\\\\r=\dfrac{a_{n+1}}{a_n}$

$a_n=400\left(\dfrac{1}{2}\right)^n\\\\a_{n+1}=400\left(\dfrac{1}{2}\right)^{n+1}\\\\r=\dfrac{a_{n+1}}{a_n}=a_{n+1}:a_n=\bigg(400\left(\frac{1}{2}\right)^{n+1}\bigg):\bigg(400\left(\frac{1}{2}\right)^n\bigg)\qquad\text{use}\ a^n:a^m=a^{n-m}\\\\=\left(\dfrac{1}{2}\right)^{n+1-n}=\left(\dfrac{1}{2}\right)^1=\dfrac{1}{2}$

Substitute:

$a_1=200,\ r=\dfrac{1}{2},\ n=6\\\\S_6=200\cdot\dfrac{1-\left(\frac{1}{2}\right)^6}{1-\frac{1}{2}}=200\cdot\dfrac{1-\frac{1}{64}}{\frac{1}{2}}=200\cdot\dfrac{63}{64}\cdot\dfrac{2}{1}=393.75$

Answer 2

Answer:   393.75

Step-by-step explanation:

$\sum\limits^6_1 {400\bigg(\dfrac{1}{2}\bigg)^n}\\\\n=1:\ 400\bigg(\dfrac{1}{2}\bigg)^1\quad =200\\n=2:\ 400\bigg(\dfrac{1}{2}\bigg)^2\quad =100\\n=3:\ 400\bigg(\dfrac{1}{2}\bigg)^3\quad =50\\n=4:\ 400\bigg(\dfrac{1}{2}\bigg)^4\quad =25\\n=5:\ 400\bigg(\dfrac{1}{2}\bigg)^5\quad =12.5\\n=6:\ 400\bigg(\dfrac{1}{2}\bigg)^6\quad =6.25\\.\qquad \qquad \qquad \qquad \quad \_\_\_\_\_\_\_\_\_\_\\\\.\qquad \qquad \quad \quad \text{Sum}=\large\boxed{393.75}$