EXAMPLE 2 Find a formula for the general term of the sequence 3 5 , − 4 25 , 5 125 , − 6 625 , 7 3125 , assuming that the patter
n of the first few terms continues. SOLUTION We are given that a1 = 3 5 a2 = − 4 25 a3 = 5 125 a4 = − 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 −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 · Incorrect: Your answer is incorrect. .
We are to find a formula for the general term of the sequence
We are given that .
The numerators of these fractions start with 3 and increase by 1. The second term has numerator 4, the third term has numerator 5; in general, the nth term will have numerator 2+n
The denominators are powers of 5, so 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 .
If I understand the drawing correctly than yes, because 6 9/12 + 6 9/12 = 12 and 18/12 = 13 and 6/12 or 13 and 1/2. 13 1/2 + 4 8/12 = 13 6/12 + 4 8/12 = 17 and 14/12 = 18 2/12 or 18 and 1/6.
Since 21 is divided by 3 you might think that it is equivalent but remeber you have to divide by both sides, so if you divide 21 by 7 it works but 28 by seven does not! Hope this helps!