What is $N \sum R = 1 {(\frac{1}{3})}^{R - 1}$ $sum_(R=1)^(N) (1/3)^(R-1)$ ? provide steps please.

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What is $N \sum R = 1 {(\frac{1}{3})}^{R - 1}$ $sum_(R=1)^(N) (1/3)^(R-1)$ ? provide steps please.

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Answer 1 · 2018-07-18T07:07:46

I got $3 / 2$ $3//2$ .

I would separate out the exponents.

$N \sum R = 1 {(\frac{1}{3})}^{R - 1}$ $sum_(R=1)^(N) (1/3)^(R-1)$

$= N \sum R = 1 {(\frac{1}{3})}^{R} {(\frac{1}{3})}^{- 1}$ $= sum_(R=1)^(N) (1/3)^R(1/3)^(-1)$

$= 3 N \sum R = 1 {(\frac{1}{3})}^{R} = 3 [{(\frac{1}{3})}^{1} + {(\frac{1}{3})}^{2} + . . .]$ $= 3sum_(R=1)^(N) (1/3)^R = 3[(1/3)^1 + (1/3)^2 + . . . ]$

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

$= 3 N \sum R = 0 {(\frac{1}{3})}^{R} - 3 {(\frac{1}{3})}^{0}$ $= 3sum_(R=0)^(N) (1/3)^R - 3(1/3)^(0)$

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

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

$3 N \sum R = 1 {(\frac{1}{3})}^{R}$ $color(blue)(3sum_(R=1)^(N) (1/3)^R)$

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

$= 3 [\frac{1}{2 / 3}] - 3$ $= 3[1/(2//3)] - 3$

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

$= \frac{3}{2}$ $= color(blue)(3/2)$

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What is $N \sum R = 1 {(\frac{1}{3})}^{R - 1}$ $sum_(R=1)^(N) (1/3)^(R-1)$ ? provide steps please.

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