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

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Answer:

Step-by-step explanation:

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

Answer: 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.

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