What is the sum of the infinite geometric series?

The answer is going to be D -6

r=1/2

Sn=a(1-r^n)/1-r

S(infinite sign) = a/1-r

S(infinite sign) = -3/1-1/2

S(infinite sign) = -3/1/2

S(infinite sign) = -3/1/1*2/2-1/2

S(infinite sign) = -3/1*2/2-1/2

S(infinite sign) = -3(3*2)

S(infinite sign) = -1*6

S(infinite sign) = -6

i dont know

step-by-step explanation:

take the logarithm of both sides (the base of the log doesn't matter):

apply the exponent property - this says that :

distribute the log terms on both sides:

group like terms together:

divide through both sides by the coefficient on :

you can do some additional rewriting here using the properties of the logarithm to simplify things if you like: