Question

What is the sum of the geometric series 2^0 + 2^1 + 2^2 + 2^3 + 2^3 + 2^4 + … + 2^9?

Answer 1

so this one you don't multiply the base by the exponent. in this case its 2 times two how ever many time the exponent says.
2^0= 2
2^1= 2
2^2=4
2^3= 8
2^4=16
2^5=32
2^6=64
2^7=128
2^8=256
2^9=512
So then you add all them up
2+2+4+8+16+32+64+128+256+512= 1024

so there is your answer 1024

Answer 2

sum is

`S_(n)=(a_(1)(1-r^(n)))/(1-r)`

r=common ratio
a1=first term
it looks like 2^0=1 is the first term aka a1
it goes to the 9th term (2^9)

sub

`S_(9)=(1(1-(2)^(9)))/(1-2)`

`S_(9)=(1-512)/(-1)`

`S_(9)=(-511)/(-1)`

`S_(9)=511`