Respuesta :
First this is not a geometric series due to the sign changes
The series can be written as A(n)=(-1)^(n-1)×2×4^(n-1)
The sum of first five will be 2-8+32-128+512=410
The series can be written as A(n)=(-1)^(n-1)×2×4^(n-1)
The sum of first five will be 2-8+32-128+512=410
now, is a geometric sequence, and sometimes called a "serie", but is just a sequence with a common "multiplier" or "common ratio".
so it goes from 2, to -8 and so on, to get the "common ratio" you just simply divide a further term by another, -8/2 = -4 <--- r
[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}\\ ----------\\ a_1=2\\ r=-4\\ n=5 \end{cases} \\\\\\ S_5=2\left( \cfrac{1-(-4)^5}{1-5} \right)\implies S_5=2\left( \cfrac{1+1024}{1-5} \right)\implies S_5=2\left( \cfrac{1025}{-4} \right)[/tex]
and surely you'd know how much that is.
so it goes from 2, to -8 and so on, to get the "common ratio" you just simply divide a further term by another, -8/2 = -4 <--- r
[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}\\ ----------\\ a_1=2\\ r=-4\\ n=5 \end{cases} \\\\\\ S_5=2\left( \cfrac{1-(-4)^5}{1-5} \right)\implies S_5=2\left( \cfrac{1+1024}{1-5} \right)\implies S_5=2\left( \cfrac{1025}{-4} \right)[/tex]
and surely you'd know how much that is.