Question

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 pattern 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. .

Answer 1

Answer:

$a_n=\dfrac{(2+n)\cdot(-1)^{n + 1}}{5^n}$

Step-by-step explanation:

We are to find a formula for the general term of the sequence

$\dfrac{3}{5}, -\dfrac{4}{25}, \dfrac{5}{125}, -\dfrac{6}{625}, \dfrac{7}{3125}$

We are given that $a_1=\dfrac{3}{5}, a_2=-\dfrac{4}{25}, a_3=\dfrac{5}{125}, a_4=-\dfrac{6}{625}, a_5=\dfrac{7}{3125}$ .

• 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 $a_n$ has denominator $5^n$
• 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 $(-1)^{n + 1}$.

Therefore,

$a_n=\dfrac{(2+n)\cdot(-1)^{n + 1}}{5^n}$