What is the sum of the geometric series in which a1 = 3, r = 4, and an = 49,152? Hint: an = a1(r)n − 1, where a1 is the first term and r is the common ratio.

Hint: cap s sub n equals start fraction a sub one left parenthesis one minus r to the power of n end power right parenthesis over one minus r end fraction comma r does not equal one comma where a1 is the first term and r is the common ratio.

Sn = −65,535
Sn = 16,383
Sn = 13,120
Sn = 65,535

i think is g(1)=-1.

hope that i can

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.

answer: you will have more water.