TLS34
Answered

what is the sum of the geometric series


15

E 2(1/4)^x

x=0


Rounded to the nearest whole number?


Sorry the format of it is odd, I'm not sure how else to type it. I don't know to solve this problem. Answer choices are:

A1
B2
C3
  1. 4

Thanks

Verified Answer :

LammettHash

If \(S\) denotes the sum of the first \(n\) terms of a geometric series with first term \(a\) and common ratio \(r\), then

\(S=a+ar+ar^2+\cdots+ar^{n-3}+ar^{n-2}+ar^{n-1}\)
\(rS=ar+ar^2+ar^3+\cdots+ar^{n-2}+ar^{n-1}+ar^n\)
\(\implies S-rS=a+(ar-ar)+(ar^2-ar^2)+\cdots+(ar^{n-1}-ar^{n-1})-ar^n\)
\(\implies (1-r)S=a(1-r^n)\)
\(\implies S=a=\dfrac{1-r^n}{1-r}\)

Using summation notation, you have

\(S=\displaystyle\sum_{x=1}^nar^{x-1}=\sum_{x=0}^{n-1}ar^x=a\dfrac{1-r^n}{1-r}\)

In this case, you have \(a=2\), \(r=\dfrac12\), and \(n=16\). So the value of the sum is

\(\displaystyle\sum_{x=0}^{15}2\left(\frac14\right)^x=2\dfrac{1-\left(\frac14\right)^{16}}{1-\frac14}\approx2.67\)

Rounded to the nearest whole number, the answer would be 3.

The sum of the considered geometric series is given by: Option C: 3

What is  a geometric sequence and how to find its nth terms?

There are three parameters which differentiate between which geometric sequence we're talking about.

  • The first parameter is the initial value of the sequence.
  • The second parameter is the quantity by which we multiply previous term to get the next term.
  • The third parameter is the length of the sequence. It can be finite or infinite.

Suppose the initial term of a geometric sequence is  \(a\) and the term by which we multiply the previous term to get the next term is  \(r\)

Then the sequence would look like

\(a, ar, ar^2, ar^3, ar^4,..\). (till the terms to which it is defined)

Thus, the nth term of such sequence would be

\(T_n = ar^{n-1}\)

What is a geometric series?

When all the terms of a geometric sequence are added, then that expression is called geometric series.

What is the sum of a geometric sequence?

Lets suppose its initial term is  \(a\) , multiplication factor is  \(r\)

and let it has total n terms, then, its sum is given as:

\(S_n = \dfrac{a(r^n-1)}{r-1}\)

(sum till nth term)

For this case, we have the terms of the considered series expressed as:

\(T_{x+1} = 2(1/4)^x\) (assuming it was for (x+1)th term

That shows that

  • a = 2,
  • r = 1/4

Now since we have:

\(\sum_{x=0}^{15} 2(1/4)^{x} = \sum_{x=0}^{15} T_{x+1} = \sum_{x=1}^{16} T_{x}\)

It is sum of starting 16 terms.

Thus, the considered series' sum for total 16 starting terms) is evaluated as:

\(S_{n=16} = \dfrac{a(r^n-1)}{r-1} = \dfrac{2((1/4)^{16} - 1)}{(1/4) - 1} \approx 2 \times \dfrac{-1}{-3/4} \approx 3\) (rounded to nearest whole number).

Thus, the sum of the considered geometric series is given by: Option C: 3

Learn more about sum of geometric series here:

https://brainly.com/question/16037289

Related Questions