Question

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

Answer 1

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

Explanation:

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

`S_n=a_1\cdot(1-r^n)/(1-r)`

We have:

`\sum\limits^6_(n=1)400\left((1)/(2)\right)^n\to a_n=400\left((1)/(2)\right)^n`

Put n = 1:

`a_1=400\left((1)/(2)\right)^1=400\left((1)/(2)\right)=200\\\\r=(a_(n+1))/(a_n)`

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

Substitute:

`a_1=200,\ r=(1)/(2),\ n=6\\\\S_6=200\cdot(1-\left((1)/(2)\right)^6)/(1-(1)/(2))=200\cdot(1-(1)/(64))/((1)/(2))=200\cdot(63)/(64)\cdot(2)/(1)=393.75`

Answer 2

Answer: 393.75

Explanation:

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