What is #sum_(R=1)^(N) (1/3)^(R-1)#? provide steps please.
1 Answer
Jul 18, 2018
I got
I would separate out the exponents.
#sum_(R=1)^(N) (1/3)^(R-1)#
#= sum_(R=1)^(N) (1/3)^R(1/3)^(-1)#
#= 3sum_(R=1)^(N) (1/3)^R = 3[(1/3)^1 + (1/3)^2 + . . . ]#
This is almost a geometric series, but it is missing the
#= 3sum_(R=0)^(N) (1/3)^R - 3(1/3)^(0)#
All we did was add and subtract
Now this is in terms of a geometric series
#color(blue)(3sum_(R=1)^(N) (1/3)^R)#
#= 3[1/(1 - (1//3))] - 3(1/3)^(0)#
#= 3[1/(2//3)] - 3#
#= 9/2 - 6/2#
#= color(blue)(3/2)#