The formula for the general term should be of the form;
#ar^(n-1)#
Where #a# is a constant, #r# is the rate at which the series is increasing, and #n# is the number of the term. The constant #a# should be the first term, in this case;
#a=4#
We can find #r# by dividing any term by the previous one. Lets use the first two.
#r=(4"/"3)/4 color(white)(color(black)("(2nd term)")/ color(black)("(1st term)")) = 1/3#
Now that we have #a# and #r# we can write our general formula.
#ar^(n-1) = 4(1/3)^(n-1)#
We should double check that we get the correct series. Plugging in the first five #n# values, we get;
#4(1/3)^(1-1), 4(1/3)^(2-1), 4(1/3)^(3-1), 4(1/3)^(4-1), 4(1/3)^(5-1)...#
#=4/3^0, 4/3^1, 4/3^2, 4/3^3, 4/3^4...#
#=4, 4/3, 4/9, 4/27, 4/81...#
Looks like it works!
