A sequence of positive real numbers a_1, a_2, a_3, \dots is defined by a_1 = 3 and a_{n+1}… — GRE General Test

A sequence of positive real numbers $a_1, a_2, a_3, \dots$ is defined by $a_1 = 3$ and $a_{n+1} = a_n + \frac{2}{a_n}$ for all integers $n \ge 1$ . Which of the following statements must be true? Select all that apply.

The sequence $a_1, a_2, a_3, \dots$ is strictly increasing.Answer
The term $a_{50}$ is strictly greater than $14$ .Answer
C
When $a_{50}$ is rounded to the nearest integer, the result is $15$ .
The sequence of consecutive term differences $d_n = a_{n+1} - a_n$ is strictly decreasing for $n \ge 1$ .Answer
E
The term $a_{100}$ is strictly greater than $25$ .

Answer

The correct statements are that the sequence is strictly increasing, the term

a_{50}

is strictly greater than 14, and the sequence of consecutive term differences

d_n = a_{n+1} - a_n

is strictly decreasing.

The statement asserting strict monotonicity is correct because

a_{n+1} - a_n = \frac{2}{a_n} > 0

for positive terms. The statement regarding

a_{50} > 14

is correct because

a_{50}^2 > 3^2 + 4(49) = 205 > 196

. The statement concerning consecutive differences is correct because

\frac{2}{a_n}

strictly decreases as

a_n

increases.

Step-by-Step Solution

Analyze monotonicity of the sequence

a_{n+1} - a_n = \frac{2}{a_n} > 0

for all

n \ge 1

Since

a_1 = 3 > 0

, all terms remain positive, making each term strictly larger than the previous.

Analyze the sequence of term differences

d_n = \frac{2}{a_n}

decreases as

a_n

increases

Since

a_n

grows strictly monotonically, its reciprocal strictly decreases, so

d_{n+1} < d_n

Derive a lower bound for

a_{50}

using quadratic expansion

a_{50}^2 > 205 \implies a_{50} > 14

Expanding

a_{k+1}^2 = a_k^2 + 4 + \frac{4}{a_k^2} > a_k^2 + 4

and summing from

k=1

49

gives

a_{50}^2 > 3^2 + 4(49) = 205 > 196 = 14^2

Derive an upper bound for

a_{50}

to test the rounding statement

a_{50} < 14.45

, so rounding to the nearest integer yields 14

Using

\frac{4}{a_k^2} < \frac{4}{4k+5}

, the sum of error terms is bounded above by

\int_0^{49} \frac{4}{4x+5} dx = \ln(40.2) \approx 3.7

. Thus

a_{50}^2 < 208.7 < 14.5^2

, so

a_{50}

rounds to 14, not 15.

Estimate

a_{100}

to test the magnitude bound

a_{100} < \sqrt{409.4} \approx 20.23 < 25

Summing the squared recurrence up to 100 terms gives

a_{100}^2 < 405 + \ln(80.2) \approx 409.4

, showing

a_{100}

cannot exceed 25.

Key Concept

Estimation of non-linear recursive sequences using squared bounds and integral comparison.

Estimated Time:3m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic