Question

Find the sum of the following infinite series $1+\frac{\sqrt{2}-1}{2\sqrt{3}}+\frac{3-2\sqrt{2}}{12}+\frac{5\sqrt{2}-7}{24\sqrt{3}}+\frac{17-12\sqrt{2}}{144}+.....$

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

Answer 1

Answer: (B)

$\frac{2\sqrt{3}}{2\sqrt{3}-\sqrt{2}+1}$

$S$ = $1+\frac{\sqrt{2}-1}{2\sqrt{3}}+\frac{3-2\sqrt{2}}{12}+\frac{5\sqrt{2}-7}{24\sqrt{3}}+\frac{17-12\sqrt{2}}{144}+.....$
${(\sqrt{2}-1)}^2=3-2\sqrt{2}$, ${(\sqrt{2}-1)}^3=5\sqrt{2}-7$
${(\sqrt{2}-1)}^4=17-12\sqrt{2}$
Hence the given series is a G.P. of infinite number of terms whose first term is $1$ and $r=\frac{\sqrt{2}-1}{2\sqrt{3}}<1$
$\therefore S_{\infty }=\frac{a}{1-r}=\frac{1}{1-\frac{\sqrt{2}-1}{2\sqrt{3}}}=\frac{2\sqrt{3}}{2\sqrt{3}-\sqrt{2}+1}$
Hence, option B.