Question

The infinite series \(3 + \frac{1}{6}x - \frac{1}{216}x^2 - \frac{1}{3888}x^3 + \cdots\) represents the Maclaurin series for the function:

• A. \(\sqrt{9 - x}\)
• B. \(\sqrt{9 + x}\)
• C. \(\sqrt{x + \frac{1}{9}}\)
• D. \(\sqrt{x - 9}\)

Hint:

Exam Tip:
For Maclaurin series recognition:

• Rewrite the function in the form \(a(1 + u)^n\).
• Compare term by term.
• Use known series expansions.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to find which function has the given Maclaurin series expansion. The Maclaurin series is a Taylor series centered at 0.

Step 2: Key Formula or Approach:
The Maclaurin series for \((1 + x)^n\) is: \[ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \cdots \] We can rewrite each option in the form \(a(1 + u)^n\) and compare coefficients.

Step 3: Detailed Explanation:
The series given is: \[ 3 + \frac{1}{6}x - \frac{1}{216}x^2 - \frac{1}{3888}x^3 + \cdots \] Factor out 3: \[ 3\left(1 + \frac{1}{18}x - \frac{1}{648}x^2 - \frac{1}{11664}x^3 + \cdots\right) \] Now, consider option (B): \(\sqrt{9 + x} = \sqrt{9} \sqrt{1 + \frac{x}{9}} = 3\left(1 + \frac{x}{9}\right)^{1/2}\).
The Maclaurin series for \((1 + u)^{1/2}\) is: \[ 1 + \frac{1}{2}u - \frac{1}{8}u^2 + \frac{1}{16}u^3 - \cdots \] Substitute \(u = \frac{x}{9}\): \[ 1 + \frac{1}{2}\left(\frac{x}{9}\right) - \frac{1}{8}\left(\frac{x}{9}\right)^2 + \frac{1}{16}\left(\frac{x}{9}\right)^3 - \cdots \] \[ = 1 + \frac{1}{18}x - \frac{1}{648}x^2 + \frac{1}{11664}x^3 - \cdots \] Multiply by 3: \[ 3 + \frac{1}{6}x - \frac{1}{216}x^2 + \frac{1}{3888}x^3 - \cdots \] The series matches the given series (except for the sign of the \(x^3\) term).
The given series has \(-\frac{1}{3888}x^3\), while the expansion gives \(+\frac{1}{3888}x^3\).
This might be due to a sign error in the question.
Option (B) is the closest match.

Step 4: Final Answer:
Therefore, option (B) is correct.