What is the sum of the first five terms of the geometric series 3 − 6 + 12 − . . . ?

Step-by-step explanation:

The nth term of a geometric sequence is expressed as;

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

a is the first term of the sequence

r is the common ratio

n is the number of terms

From the sequence;

a = 3

r = -6/3 = 12/-6 = -2

n = 5 (since are looking for the sum of first five terms)

Substitute into the formula;

S5 = 3((-2)^5 - 1)/-2 - 1

S5 = 3(-32-1)/-3

S5 = 3(-33)/-3

S5 = -(-33)

S5 = 33

Hence the sum of the first five terms of the sequence is 33

10

step-by-step explanation:

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation: