Answer:
2,046
Step-by-step explanation:
[tex]\sum\limits_{k=1}^{5}6\left(4\right)^{k-1}=6\left(4\right)^{1-1}+6\left(4\right)^{2-1}+6\left(4\right)^{3-1}+6\left(4\right)^{4-1}+6\left(4\right)^{5-1}=\\\\\\=6\cdot4^0+6\cdot4^1+6\cdot4^2+6\cdot4^3+6\cdot4^4=6+24+96+384+1536=\\\\=2,046[/tex]
or:
is geometric series: [tex]S_n=\dfrac{a_1(1-q^n)}{1-q}[/tex]
[tex]a_n=a_1\cdot q^{n-1}\quad\implies\quad\begin{cases}a_1=6\\q=4\\n=5\end{cases}[/tex]
so:
[tex]\sum\limits_{k=1}^{5}6\left(4\right)^{k-1}=S_5=\dfrac{6(1-4^5)}{1-4}=\dfrac{6(1-1,024)}{-3}=-2(-1,023)=2,046[/tex]
