Question

Find the sum of the following series (with infinite terms):
\(2\sqrt2, \frac4{\sqrt3}, \frac{4\sqrt2}{3}, .....\)

• A. \(2\sqrt3 (\sqrt3+\sqrt2)\)
• B. \(2\sqrt3 (\sqrt3-\sqrt2)\)
• C. \(2\sqrt6 (\sqrt3+\sqrt2)\)
• D. \(2\sqrt6 (\sqrt3-\sqrt2)\)

Answer 1

The Correct Option is C

To find the sum of the given infinite series, we first need to identify the pattern and determine if it forms a geometric progression.

The series given is: \(2\sqrt{2}, \frac{4}{\sqrt{3}}, \frac{4\sqrt{2}}{3}, \ldots\)

Let's denote the first term as \( a_1 = 2\sqrt{2} \) and calculate the common ratio \( r \) by dividing the second term by the first term.

Second term: \( \frac{4}{\sqrt{3}} \)

Third term: \( \frac{4\sqrt{2}}{3} \)

The common ratio \( r \) is calculated as follows:

\( r = \frac{\text{second term}}{\text{first term}} = \frac{\frac{4}{\sqrt{3}}}{2\sqrt{2}} = \frac{4}{2\sqrt{2}\sqrt{3}} = \frac{4}{2\sqrt{6}} = \frac{2}{\sqrt{6}} \)

Check the ratio for the third term to ensure it remains consistent:

\( r = \frac{\text{third term}}{\text{second term}} = \frac{\frac{4\sqrt{2}}{3}}{\frac{4}{\sqrt{3}}} = \frac{4\sqrt{2}}{3} \cdot \frac{\sqrt{3}}{4} = \frac{\sqrt{6}}{3} \)

After simplifying, we observe that both calculations for \( r \) return the same value, confirming that the series forms a geometric progression with:

• First term \( a = 2\sqrt{2} \)
• Common ratio \( r = \frac{\sqrt{6}}{3} \)

The sum of an infinite geometric series is given by the formula:

\( S = \frac{a}{1 - r} \)

Substitute the values of \( a \) and \( r \) into the formula:

\( S = \frac{2\sqrt{2}}{1 - \frac{\sqrt{6}}{3}} = \frac{2\sqrt{2}}{\frac{3 - \sqrt{6}}{3}} = \frac{2\sqrt{2} \times 3}{3 - \sqrt{6}} \)

Simplify further by rationalizing the denominator:

Multiply numerator and denominator by the conjugate of the denominator:

\( S = \frac{2\sqrt{2} \times 3 \times (3 + \sqrt{6})}{(3 - \sqrt{6})(3 + \sqrt{6})} = \frac{6\sqrt{2}(3 + \sqrt{6})}{9 - 6} = \frac{6\sqrt{2}(3 + \sqrt{6})}{3} \)

Finally, simplify the expression:

\( S = 2\sqrt{2} (3 + \sqrt{6}) = 2\sqrt{2} \times 3 + 2\sqrt{2} \times \sqrt{6} = 6\sqrt{2} + 2\sqrt{12} = 6\sqrt{2} + 4\sqrt{3} \)

Therefore, the sum of the infinite series is correctly given by the answer \(2\sqrt{6}(\sqrt{3}+\sqrt{2})\).