Question

The sum of the series $(\sqrt{2}+1)+1+(\sqrt{2}-1)+\cdots$ up to infinite term is

• A. $\frac{1+3\sqrt{2}}{2}$
• B. $\frac{3+\sqrt{2}}{2}$
• C. $\frac{5+2\sqrt{2}}{2}$
• D. $\frac{4+3\sqrt{2}}{2}$

Answer 1

The correct option is D $\frac{4+3\sqrt{2}}{2}$

Let $S=(\sqrt{2}+1)+1+(\sqrt{2}-1)+\cdots \cdots$
$\Rightarrow \frac{S}{\sqrt{2}+1}=1+\frac{1}{\sqrt{2}+1}+\frac{\sqrt{2}-1}{\sqrt{2}+1}+\cdots \Rightarrow \frac{S}{\sqrt{2}+1}=1+\frac{1}{\sqrt{2}+1}+\frac{1}{(\sqrt{2}+1{)}^2}+\cdots \Rightarrow \frac{S}{\sqrt{2}+1}=\frac{1}{1-\frac{1}{(\sqrt{2}+1)}}\Rightarrow \frac{S}{\sqrt{2}+1}=\frac{\sqrt{2}+1}{\sqrt{2}}\Rightarrow S=\frac{3+2\sqrt{2}}{\sqrt{2}}\therefore S=\frac{4+3\sqrt{2}}{2}$