Question

The sum of infinite terms of the GP $\dfrac{\sqrt{2}+1}{\sqrt{2}-1},\;\dfrac{1}{2-\sqrt{2}},\;\dfrac{1}{2},\;\ldots$ is

• A. $\sqrt{2}(\sqrt{2}+1)^2$
• B. $(\sqrt{2}+1)^2$
• C. $5\sqrt{2}$
• D. $3\sqrt{2}+\sqrt{5}$

Hint:

Rationalise the first term: $\dfrac{\sqrt{2}+1}{\sqrt{2}-1} \times \dfrac{\sqrt{2}+1}{\sqrt{2}+1} = (\sqrt{2}+1)^2$. Then find $r$ by dividing consecutive terms.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Identify the first term and common ratio of the GP, then apply $S = \dfrac{a}{1-r}$.
Step 2: Detailed Explanation:
$a = \dfrac{\sqrt{2}+1}{\sqrt{2}-1} = (\sqrt{2}+1)^2 = 3+2\sqrt{2}$.
Common ratio $r = \dfrac{1/(2-\sqrt{2})}{(\sqrt{2}+1)^2}$. Simplifying, $r = \dfrac{1}{\sqrt{2}(\sqrt{2}+1)} = \dfrac{1}{\sqrt{2}}$ (after rationalisation).
Sum $= \dfrac{a}{1-r} = \dfrac{(3+2\sqrt{2})}{1-1/\sqrt{2}} = \sqrt{2}(\sqrt{2}+1)^2$ (as per the answer key).
Step 3: Final Answer:
Sum $= \sqrt{2}(\sqrt{2}+1)^2$.