What is the sum of the series $$(-1/3)^0+(-1/3)^1+(-1/3)^2+(-1/3)^3+\ldots$$ Remember to enter your answer in lowest terms.

Question

BlackTom AI · Accepted Answer

We are given the infinite series (-1/3)^0 + (-1/3)^1 + (-1/3)^2 + (-1/3)^3 + ...

First, recognize this as a geometric series with first term a = 1 (since (-1/3)^0 = 1) and common ratio r = -1/3.

Since |r| = 1/3 < 1, the infinite sum converges. The standard formula for the sum of a geometric series is S = a / (1 - r). Here, that gives S = 1 / (1 - (-1/3)) = 1 / (1 + 1/3) = 1 / (4/3) = 3/4.

An alternative way to see this is to use the partial-sum formula S_n = (1 - r^{n+1}) / (1 - r) and note that as n → ∞, r^{n+1} → 0 because |r| < 1. This yields the same limit S = 1 / (1 - r) = 3/4.

The computed sum 3/4 is already in lowest terms. There are no answer choices to evaluate here, but the result is 3/4.

What is the sum of the series \[(-1/3)^0+(-1/3)^1+(-1/3)^2+(-1/3)^3+\ldots\] Remember to enter your answer in lowest terms.简答题

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What is the sum of the series \[(-1/3)^0+(-1/3)^1+(-1/3)^2+(-1/3)^3+\ldots\] Remember to enter your answer in lowest terms.简答题

类似问题

Which one of the following is a geometric series?

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What would be the sum of the first 18 terms of the geometric series with: 20 1.4 (type answer only in box correct to the nearest whole number)

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希望你的学习变得更简单

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