Question

Real quick help please:
What is the sum of the geometric series 15Σx=0 2(1/4)^x rounded to the nearest whole number?

Answer 1

\[ 15 \sum_{x=0}^2\left(\frac{1}{4}\right)^x \,\, \text{?}\]

Answer 2

Yes, with a 2 in front of 1/4

Answer 3

and the 15 is on top of the sigma

Answer 4

\[\sum_{x=0}^{15}2\left(\frac{1}{4}\right)^x \]

Answer 5

yes, that is perfect.

Answer 6

My options are 1, 2, 3, and 4.

Answer 7

It's 3 to the nearest integer

Answer 8

It is?

Answer 9

Can you show me how to figure that out?

Answer 10

There is a formula for geometric series:
\[ \large \sum_{x=0}^n ar^x=a\frac{1-r^{n+1}}{1-r}\]. Here, \(a\) is any number (n your example it is 2), and \(r\) must be between 0 and 1, and here \(r=1/4\)

Answer 11

then you calculate \[2\cdot \frac{1-(\frac{1}{4})^{16}}{1-\frac{1}{4}} \]

Answer 12

okay
Where does the 16th power on 1/4 come from?

Answer 13

Your summation goes from 0 to 15. The formula says you have "0 to n", so n=15. But look in the formula you have \(1-r^{n+1}\), so you have \(1-r^{15+1}\)

Answer 14

Oh and I said r should be between 0 and 1, but actually it's valid for |r| <1