Respuesta :
If [tex]S[/tex] denotes the sum of the first [tex]n[/tex] terms of a geometric series with first term [tex]a[/tex] and common ratio [tex]r[/tex], then
[tex]S=a+ar+ar^2+\cdots+ar^{n-3}+ar^{n-2}+ar^{n-1}[/tex]
[tex]rS=ar+ar^2+ar^3+\cdots+ar^{n-2}+ar^{n-1}+ar^n[/tex]
[tex]\implies S-rS=a+(ar-ar)+(ar^2-ar^2)+\cdots+(ar^{n-1}-ar^{n-1})-ar^n[/tex]
[tex]\implies (1-r)S=a(1-r^n)[/tex]
[tex]\implies S=a=\dfrac{1-r^n}{1-r}[/tex]
Using summation notation, you have
[tex]S=\displaystyle\sum_{x=1}^nar^{x-1}=\sum_{x=0}^{n-1}ar^x=a\dfrac{1-r^n}{1-r}[/tex]
In this case, you have [tex]a=2[/tex], [tex]r=\dfrac12[/tex], and [tex]n=16[/tex]. So the value of the sum is
[tex]\displaystyle\sum_{x=0}^{15}2\left(\frac14\right)^x=2\dfrac{1-\left(\frac14\right)^{16}}{1-\frac14}\approx2.67[/tex]
Rounded to the nearest whole number, the answer would be 3.
[tex]S=a+ar+ar^2+\cdots+ar^{n-3}+ar^{n-2}+ar^{n-1}[/tex]
[tex]rS=ar+ar^2+ar^3+\cdots+ar^{n-2}+ar^{n-1}+ar^n[/tex]
[tex]\implies S-rS=a+(ar-ar)+(ar^2-ar^2)+\cdots+(ar^{n-1}-ar^{n-1})-ar^n[/tex]
[tex]\implies (1-r)S=a(1-r^n)[/tex]
[tex]\implies S=a=\dfrac{1-r^n}{1-r}[/tex]
Using summation notation, you have
[tex]S=\displaystyle\sum_{x=1}^nar^{x-1}=\sum_{x=0}^{n-1}ar^x=a\dfrac{1-r^n}{1-r}[/tex]
In this case, you have [tex]a=2[/tex], [tex]r=\dfrac12[/tex], and [tex]n=16[/tex]. So the value of the sum is
[tex]\displaystyle\sum_{x=0}^{15}2\left(\frac14\right)^x=2\dfrac{1-\left(\frac14\right)^{16}}{1-\frac14}\approx2.67[/tex]
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 [tex]a[/tex] and the term by which we multiply the previous term to get the next term is [tex]r[/tex]
Then the sequence would look like
[tex]a, ar, ar^2, ar^3, ar^4,..[/tex]. (till the terms to which it is defined)
Thus, the nth term of such sequence would be
[tex]T_n = ar^{n-1}[/tex]
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 [tex]a[/tex] , multiplication factor is [tex]r[/tex]
and let it has total n terms, then, its sum is given as:
[tex]S_n = \dfrac{a(r^n-1)}{r-1}[/tex]
(sum till nth term)
For this case, we have the terms of the considered series expressed as:
[tex]T_{x+1} = 2(1/4)^x[/tex] (assuming it was for (x+1)th term
That shows that
- a = 2,
- r = 1/4
Now since we have:
[tex]\sum_{x=0}^{15} 2(1/4)^{x} = \sum_{x=0}^{15} T_{x+1} = \sum_{x=1}^{16} T_{x}[/tex]
It is sum of starting 16 terms.
Thus, the considered series' sum for total 16 starting terms) is evaluated as:
[tex]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[/tex] (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:
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