Answer:
[tex]4095[/tex]
Step-by-step explanation:
[tex]\sum _{n=1}^63\left(4^{n-1}\right)\:[/tex]
[tex]\sum _{n=1}^63\cdot \:4^{n-1}[/tex]
Compute general progression:
[tex]\frac{3\cdot \:4^{\left(n+1\right)-1}}{3\cdot \:4^{n-1}}=4[/tex]
[tex]r=4\\[/tex]
[tex]a_1=3\cdot \:4^{1-1}[/tex]
[tex]a_1=3[/tex]
nth term is computed by:
[tex]r=4,\:a_n=3\cdot \:4^{n-1}[/tex]
plug in the values [tex]n=6,\:\spacea_1=3,\:\spacer=4[/tex]:
[tex]=3\cdot \frac{1-4^6}{1-4}[/tex]
[tex]=4095[/tex]
