Question

How do you evaluate the sum of the finite geometric series: 1+4+16+...+4096?

Answer 1

\[a=1,r=\frac{ 4 }{ 1 }=4,t _{n}=ar ^{n-1}\]
\[4096=4^{n-1}\]

\[4^{n-1}=4^6,n-1=6,n=6+1=7\]
\[S _{n}=\frac{ a \left( r^n-1 \right) }{ r-1 }\]
now you can calculate.

Answer 2

so n is 7?

Answer 3

correct.

Answer 4

hint\[4^7=4^6*4\]

Answer 5

so \[4096 = \frac{ 1-4^7 }{ 1-4 }\] correct?

Answer 6

\[S _{7}=\frac{ 1\left( 4^7-1 \right) }{ 4-1 }=\frac{ 4^6*4-1 }{ 3 }=\frac{ 4096*4-1 }{ 3 }=\frac{ 16384-1 }{ 3 }\]
=?