Write the sum using summation notation, assuming the suggested pattern continues.

1 - 3 + 9 - 27 + ...

A. summation of one times three to the power of n from n equals zero to infinity
B. summation of one times three to the power of the quantity n plus one from n equals zero to infinity
C. summation of one times negative three to the power of n from n equals zero to infinity
D. summation of one times negative three to the power of the quantity n plus one from n equals zero to infinity

WILL MARK BRAINLIEST IF ANSWERED IN 15 MINUTES

Step-by-step explanation:

It is upper limit infinity lower limit n = 0 then you would put the stigma sign 1(-3)^n

B.

Step-by-step explanation:

1 - 3 + 9 - 27 +...

So first you would need to make that one to an equvalinet number, which is 100.

So 100 - 3 = 97 .

97 + 9 = 106.

106 - 27 = 79.

So u would need to do 1 x 3 = 3

So the correct answer is...

1 - 3 + 9 - 27 + 3.

Your welcome!

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8