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