Mathematics, 07.04.2020 16:24 asclp
EXAMPLE 2 Find a formula for the general term of the sequence 3 5 , β 4 25 , 5 125 , β 6 625 , 7 3125 , assuming that the pattern of the first few terms continues. SOLUTION We are given that a1 = 3 5 a2 = β 4 25 a3 = 5 125 a4 = β 6 625 a5 = 7 3125 . Notice that the numerators of these fractions start with 3 and increase by 1 Correct: Your answer is correct. whenever we go to the next term. The second term has numerator 4, the third term has numerator 5; in general, the nth term will have numerator Incorrect: Your answer is incorrect. . The denominators are powers of 5 Correct: Your answer is correct. , so an has denominator Correct: Your answer is correct. . The signs of the terms are alternately positive and negative so we need to multiply by a power of β1. Here we want to start with a positive term and so we use (β1)n β 1 or (β1)n + 1. Therefore, an = (β1)n β 1 Β· Incorrect: Your answer is incorrect. .
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EXAMPLE 2 Find a formula for the general term of the sequence 3 5 , β 4 25 , 5 125 , β 6 625 , 7 312...
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