Respuesta :
[tex]\bf \begin{array}{|cc|ll} \cline{1-2} n&a_n\\ \cline{1-2} 1&4\\ &\\ 2&\stackrel{4(-3)}{-12}\\ &\\ 3&\stackrel{-12(-3)}{36}\\ \cline{1-2} \end{array}\qquad \impliedby \textit{common ratio of "r" is -3} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\[-0.5em] \hrulefill\\ r=-3\\ a_1=4\\ n=15 \end{cases} \\\\\\ S_{15}\implies \displaystyle\sum\limits_{i=4}^{15}~4(-3)^{i-1}[/tex]
Answer:
[tex]\sum_{n=4}^{15}4(-3)^{n-1}[/tex]
Step-by-step explanation:
The given sequence is 4, -12, 36
We can see there is a common ratio
[tex]\frac{a_{2} }{a_{1} }[/tex] = [tex]\frac{-12}{4}=(-3)[/tex]
[tex]\frac{a_{2} }{a_{3} }[/tex] = [tex]\frac{36}{-12}=(-3)[/tex]
Therefore, the given sequence is a geometric sequence.
Now we have to determine the sigma notation of the sum for term 4 through term 15.
Since explicit formula of the sigma can be represented as
[tex]T_{n}=a(r)^{n-1}[/tex]
where [tex]T_{n}[/tex] = nth term
a = first term
n = number of term term
r = common ratio
and sum is denoted by [tex]\sum_{n=1}^{n}a(r)^{n-1}[/tex]
Now for the given sequence sigma notation will be
[tex]\sum_{n=4}^{15}4(-3)^{n-1}[/tex]
