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

Jul 18, 2018

I got $3 / 2$.

I would separate out the exponents.

${\sum}_{R = 1}^{N} {\left(\frac{1}{3}\right)}^{R - 1}$

$= {\sum}_{R = 1}^{N} {\left(\frac{1}{3}\right)}^{R} {\left(\frac{1}{3}\right)}^{- 1}$

$= 3 {\sum}_{R = 1}^{N} {\left(\frac{1}{3}\right)}^{R} = 3 \left[{\left(\frac{1}{3}\right)}^{1} + {\left(\frac{1}{3}\right)}^{2} + . . .\right]$

This is almost a geometric series, but it is missing the $R = 0$ term. Therefore, we rewrite this as:

$= 3 {\sum}_{R = 0}^{N} {\left(\frac{1}{3}\right)}^{R} - 3 {\left(\frac{1}{3}\right)}^{0}$

All we did was add and subtract $3 {\left(\frac{1}{3}\right)}^{0}$.

Now this is in terms of a geometric series ${\sum}_{R = 0}^{N} {r}^{R}$, and since $0 < r < 1$, this converges as

$\textcolor{b l u e}{3 {\sum}_{R = 1}^{N} {\left(\frac{1}{3}\right)}^{R}}$

$= 3 \left[\frac{1}{1 - \left(1 / 3\right)}\right] - 3 {\left(\frac{1}{3}\right)}^{0}$

$= 3 \left[\frac{1}{2 / 3}\right] - 3$

$= \frac{9}{2} - \frac{6}{2}$

$= \textcolor{b l u e}{\frac{3}{2}}$