Question

Consider the geometric seriesΣ[infinity]ₙ=1/12ⁿ . if the sum is a perfect square, what is the smallest possible value of n, where is a positive integer?

A) 2
B) 3
C) 4
D) 5

Answer 1

To find the smallest possible value of n where the sum of the geometric series Σ∞ₙ=1/12ⁿ is a perfect square, we can use the formula for the sum of a geometric series. Therefore, the answer is D) 5.

Step-by-step explanation:

To find the smallest possible value of n where the sum of the geometric series Σ∞ₙ=1/12ⁿ is a perfect square, we can use the formula for the sum of a geometric series. The formula is: S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio. In this case, a = 1/12 and r = 1/12. So the sum is:

S = (1/12) / (1 - 1/12) = (1/12) / (11/12) = 1/11

Since 1/11 is not a perfect square, we need to find the next term of the series. The next term is 1/12² = 1/144. So the new sum is:

S = (1/12 + 1/144) / (1 - 1/12) = (1/12 + 1/144) / (11/12) = 13/132

Since 13/132 is still not a perfect square, we continue adding terms until we get a perfect square. The next term is 1/12³ = 1/1728. So the new sum is:

S = (1/12 + 1/144 + 1/1728) / (1 - 1/12) = (1/12 + 1/144 + 1/1728) / (11/12) = 541/3564

541/3564 is still not a perfect square, so we continue adding terms. The next term is 1/12⁴ = 1/20736. So the new sum is:

S = (1/12 + 1/144 + 1/1728 + 1/20736) / (1 - 1/12) = (1/12 + 1/144 + 1/1728 + 1/20736) / (11/12) = 4681/30128

4681/30128 is still not a perfect square. We continue this process until we get a perfect square. The smallest value of n where the sum of the geometric series is a perfect square is n = 5. Therefore, the answer is D) 5.