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Consider the series

∑ₙ=1[infinity](7²-4²)(4/7)ⁿ
This series is a geometric series. What is its limit?

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Final answer:

The limit of the given geometric series is 77.

Step-by-step explanation:

The given series, ∑ₙ=1[infinity](7²-4²)(4/7)ⁿ, is indeed a geometric series. A geometric series is a sequence of numbers in which each term is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio. In this case, the common ratio is (4/7). To find the limit of the series, we can use the formula for the sum of an infinite geometric series:

S = a/(1 - r)

where S is the sum of the series, a is the first term, and r is the common ratio. The first term of the series is (7²-4²) = 33, so substituting the values into the formula:

S = 33/(1 - 4/7)

Simplifying the expression:

S = 33/(3/7)

S = (33 * 7)/3

S = 231/3

S = 77

Therefore, the limit of the given geometric series is 77.

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