Question

For the following sequence in G.P., find the sum of infinite terms. (√2+1)/(√2-1) + 1/(2-√2) + 1/2 + .........​

Answer 1

The sum of infinite terms is $\sqrt{2}(\sqrt{2}+1)^{2}$

Explanation:

The series is $\frac{\sqrt{2}+1}{\sqrt{2}-1}, \frac{1}{2-\sqrt{2}}, \frac{1}{2} ,.....$

The first term is $a=\frac{\sqrt{2}+1}{\sqrt{2-1}}$

Taking conjugate, we get,

$\begin{aligned}a &=\frac{\sqrt{2}+1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \\&=\frac{(\sqrt{2}+1)^{2}}{2-1} \\&=\frac{2+1+2 \sqrt{2}}{1} \\&=3+2 \sqrt{2}\end{aligned}$

Now, we shall determine the common difference,

$r=\frac{\frac{1}{2-\sqrt{2}}}{\frac{\sqrt{2}+1}{\sqrt{2}-1}}$

Simplifying,

$\begin{aligned}r &=\frac{1}{\sqrt{2}(\sqrt{2}-1)} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} \\&=\frac{\sqrt{2}-1}{\sqrt{2}}\end{aligned}$

The sum of infinite terms can be determined using the formula,

$S_{\infty}=\frac{a}{1-r}$

Substituting the values, we get,

$\begin{aligned}S_{\infty} &=\frac{3+2 \sqrt{2}}{1-\frac{\sqrt{2}-1}{\sqrt{2}}} \\&=\frac{\sqrt{2}(3+2 \sqrt{2})}{1} \\&=\sqrt{2}(2+1+2 \sqrt{2}) \\&=\sqrt{2}(\sqrt{2}+1)^{2}\end{aligned}$

Thus, the sum of infinite terms is $\sqrt{2}(\sqrt{2}+1)^{2}$

Learn more:

1. brainly.in/question/12426103

2. brainly.in/question/1168520