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

$color(blue)(3sum_(R=1)^(N) (1/3)^R)$

$= 3[1/(1 - (1//3))] - 3(1/3)^(0)$

$= 3[1/(2//3)] - 3$

$= 9/2 - 6/2$

$= color(blue)(3/2)$

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