What is #sum_(R=1)^(N) (1/3)^(R-1)#? provide steps please.

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Answer 1 · 2018-07-18T07:07:46

I got #3//2#.

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 #R = 0# term. Therefore, we rewrite this as:

#= 3sum_(R=0)^(N) (1/3)^R - 3(1/3)^(0)#

All we did was add and subtract #3(1/3)^(0)#.

Now this is in terms of a geometric series #sum_(R=0)^(N) r^R#, and since #0 < r < 1#, this converges as

#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)#