Question

The value of $\sqrt{1+2\sqrt{1+2\sqrt{1+2+....}}}$ is

• A. $2+\sqrt{2}$
• B. $1+\sqrt{2}$
• C. $3+\sqrt{2}$
• D. $3+\sqrt{3}$

Answer 1

Answer: (B)

$1+\sqrt{2}$

$y=\sqrt{1+2\sqrt{1+2\sqrt{1+2+}}}\dots \dots \dots \infty y=\sqrt{1+2y}⟹y^2-2y-1=0y=\frac{2\pm \sqrt{4-4(-1)}}{2}=\frac{2\pm 2\sqrt{2}}{2}=1+\sqrt{2}$

Hence, this is the answer.