Find a formula for the general term of the sequence 4 5 6 3725'...} 25 125 625 3125 assuming that the pattern of the first few terms continues. EXAMPLE 2 3 I - SOLUTION We are given that 3 a ==// a2 = 5 of 5 7 an = 4 25 (1)n-1. a3 = 5 125 X Notice that the numerators of these fractions start with 3 and increase by whenever we go to the next term. The second term has numerator 4, the third term has numerator 5; in general, the nth term 1 will have numerator (n+1) 5n a4 . = 6 625 a5 = 7 3125 , so an has denominator 5 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, x ). The denominators are powers The signs of