# The partial sum -3+(-9)+(-27)+...+(-2187) equals?

### 5 Answers

- husoskiLv 71 month agoFavorite Answer
That a geometric series with n=7 terms, first term a=-3 and common ratio r=3.

The sum is a finite geometric series is well known:

a + ar + ar^2 + ... + ar^(n-1) = a (1 - r^n) / (1 - r)

= (-3) * (1 - 3^7) / (1 - 3)

I get -3279 on my calculator.

- sepiaLv 71 month ago
-3 + (-9) + (-27) + ... + (-2187)

sum_(n = 1)^7 - 3^n = -3279

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- PhilipLv 61 month ago
Partial sum, PS, = -[3+3*3+3*3^2+3*3^3+,...3*3^6]. General form of a geometric

progression, (GP), is a,ar,ar^2,ar^3,....,ar^(n-1). Then -PS is a truncated GP with a =

3, r = 3 and n = 7. Sum of the 1st n terms of the general GP = S(n) = [(r^n-1)/(r-1)]a,

where r =/= 1. Then PS = -S(7) = -[{3^(7-1) -1}/(3-1)](3) = -(3/2)(3^6 -1) = -1092.

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thanks!

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- llafferLv 71 month ago
This is the sum of a geometric sequence where the first term is -3 and the common ratio is 3.

I use this expression to describe the n'th term of a geometric sequence:

a(n) = a r^(n - 1)

We know "a" and "r". if we then set a(n) to -2187 we can solve for n to find out how many terms are in this sequence:

a(n) = -3(3)^(n - 1)

-2187 = -3(3)^(n - 1)

729 = 3^(n - 1)

3^6 = 3^(n - 1)

6 = n - 1

n = 7

-2187 is the 7th term.

Now that we know that, we can use this equation which gives us the sum of the first "n" terms of a geometric sequence:

S(n) = a(1 - r^n) / (1 - r)

Using a = -3, r = 3, and n = 7, solve for S(7):

S(n) = -3(1 - 3^n) / (1 - 3)

S(n) = -3(1 - 3^n) / -2

S(n) = (3/2)(1 - 3^n)

S(7) = (3/2)(1 - 3^7)

S(7) = (3/2)(1 - 2187)

S(7) = (3/2)(-2186)

S(7) = -3279

The sum of your series is -3279

thanks!

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thanks!