Question

Please help me I’m so lost !!!!

Answer 1

Answer:

$4095$

Step-by-step explanation:

$\sum _{n=1}^63\left(4^{n-1}\right)\:$

$\sum _{n=1}^63\cdot \:4^{n-1}$

Compute general progression:

$\frac{3\cdot \:4^{\left(n+1\right)-1}}{3\cdot \:4^{n-1}}=4$

$r=4$

$a_1=3\cdot \:4^{1-1}$

$a_1=3$

nth term is computed by:

$r=4,\:a_n=3\cdot \:4^{n-1}$

plug in the values $n=6,\:\spacea_1=3,\:\spacer=4$:

$=3\cdot \frac{1-4^6}{1-4}$

$=4095$