1/16
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The above answer of 1/16 is the answer your teacher is more likely wanting as it makes the series so far a GP (Geometric Progression) with U{n} = 64 × (¼)ⁿ⁻¹. However, it is possible to find infinitely many polynomials which also give {64, 16, 4, 1, ¼} for the first five terms, but then diverge and continue the sequence in different ways, for example:
U{n} = (-153n⁵ + 3740n⁴ - 32735n³ + 131560n² - 246172n + 174480)/480
gives {64, 16, 4, 1, ¼} for n = {1, 2, 3, 4, 5}, so the next term U{6} = 42
U{n} = (27n⁴ - 414n³ + 2385n² - 6198n + 6248)/32
gives {64, 16, 4, 1, ¼} for n = {1, 2, 3, 4, 5}, so the next term U{6} = 15¼
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