A sequence of numbers a_1, a_2, a_3, \dots is defined by a_1 = 5 and a_{n+1} = a_n^2 - 2 f… — GRE General Test

A sequence of numbers $a_1, a_2, a_3, \dots$ is defined by $a_1 = 5$ and $a_{n+1} = a_n^2 - 2$ for all positive integers $n \ge 1$ . Which of the following is the value of the sum $\sum_{n=1}^{4} \frac{1}{a_n}$ , rounded to the nearest hundredth?

A
0.20
B
0.24
0.25Answer
D
0.32
E
0.43

Answer

0.25

Evaluating the recurrence relation yields

a_1 = 5

a_2 = 23

a_3 = 527

, and

a_4 = 277,727

. Summing their reciprocals produces

\frac{1}{5} + \frac{1}{23} + \frac{1}{527} + \frac{1}{277,727} \approx 0.20 + 0.043478 + 0.001898 + 0.000004 = 0.24538

. Rounding

0.24538

to the nearest hundredth yields 0.25.

Step-by-Step Solution

Calculate the first four terms of the defined sequence using the recurrence relation

a_{n+1} = a_n^2 - 2

a_1 = 5

a_2 = 5^2 - 2 = 23

a_3 = 23^2 - 2 = 527

, and

a_4 = 527^2 - 2 = 277,727

The recurrence rule determines each subsequent term from the preceding term.

Compute the sum of reciprocals

\sum_{n=1}^{4} \frac{1}{a_n} = \frac{1}{5} + \frac{1}{23} + \frac{1}{527} + \frac{1}{277,727}

\frac{1}{5} = 0.2

\frac{1}{23} \approx 0.043478

\frac{1}{527} \approx 0.001898

, and

\frac{1}{277,727} \approx 0.0000036

. Sum

\approx 0.24538

Converting each fraction term to decimal form allows straightforward addition.

Round the calculated sum

0.24538

to the nearest hundredth.

Since the thousandths digit is

5

0.24538

rounds up to

0.25

Standard rounding rules dictate rounding up when the digit to the right of the target decimal place is

5

or greater.

Key Concept

Defined sequence terms evaluation, reciprocal sum estimation, and decimal rounding

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic