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These are the first six terms of a sequence with a₁=28 : 28,21,(63)/(4),(189)/(16),(567)/(64),(1701)/(256),dots

Find a recursive formula for this sequence that is valid for n>1.

User Brydenr
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1 Answer

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Final answer:

The sequence can be described by a recursive formula where each term is the product of the previous term and 3/4. Thus, the recursive formula is a(n) = (3/4) * a(n - 1) for n > 1 with a(1) = 28.

Step-by-step explanation:

To find a recursive formula for the given sequence 28, 21, 63/4, 189/16, 567/64, 1701/256,... we observe the pattern of both the numerators and the denominators. The numerators (28, 21, 63, 189, 567, 1701,...) form a sequence where each term, after the first, is the product of the previous term and 3/4. Similarly, the denominators (1, 4, 16, 64, 256,...) are powers of 4 (4^0, 4^1, 4^2, 4^3,...).

The formula for the nth term of a geometric sequence is a(n) = a(1) * r^(n-1), where a(1) is the first term and r is the common ratio. Our sequence can be thought of as a product of two geometric sequences, one for the numerator with a common ratio of 3/4, and one for the denominator with a common ratio of 4.

Hence, the recursive formula for n > 1 can be defined as:

  • a(n) = (3/4) * a(n - 1) where n > 1 and a(1) = 28.
User Glls
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