Question

Prove the following relation ship

$\sf \: \sqrt{2 + \sqrt{2 + \sqrt{2 + - - - + \sqrt{2} } } } = 2cos( \frac{\pi}{ {2}^{n + 1} } ) \\ \\ \sf \: where \: root \: repeat \: n \: times \: and \: n \: is \: natural \: number$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sqrt{2 + \sqrt{2 + \sqrt{2 + - - - + \sqrt{2} } } } = 2cos\bigg( \dfrac{\pi}{ {2}^{n + 1} } \bigg)$

To prove this result, we use Principal of Mathematical Induction.

Let assume that

$\rm \:P(n) : \sqrt{2 + \sqrt{2 + \sqrt{2 + - - - + \sqrt{2} } } } = 2cos\bigg( \dfrac{\pi}{ {2}^{n + 1} } \bigg)$

Step :- 1 For n = 1

$\rm \:P(1) : \sqrt{2} = 2cos\bigg( \dfrac{\pi}{ {2}^{1 + 1} } \bigg)$

$\rm \:P(1) : \sqrt{2} = 2cos\bigg( \dfrac{\pi}{ {2}^{2} } \bigg)$

$\rm \:P(1) : \sqrt{2} = 2cos\bigg( \dfrac{\pi}{4} \bigg)$

$\rm \:P(1) : \sqrt{2} = 2 \times \frac{1}{ \sqrt{2} }$

$\rm \:P(1) : \sqrt{2} = \sqrt{2}$

$\rm\implies \:P(n) \: is \: true \: for \: n = 1$

Step :- 2 Let assume that P(n) is true for n = k, where k is some natural number.

So,

$\rm \:P(k) : \sqrt{2 + \sqrt{2 + \sqrt{2 + - - - + \sqrt{2} } } } = 2cos\bigg( \dfrac{\pi}{ {2}^{k + 1} } \bigg)$

Step :- 3 Now, we have to prove that P(n) is true for n = k + 1.

It means, we have to prove that

$\rm \:P(k + 1) : \underbrace{\sqrt{2 + \sqrt{2 + \sqrt{2 + - - - + \sqrt{2} } } } }= 2cos\bigg( \dfrac{\pi}{ {2}^{k + 2} } \bigg) \\ \: \: \: \: (k + 1) \: terms$

Now, Consider LHS

$\rm \: \underbrace{\sqrt{2 + \sqrt{2 + \sqrt{2 + - - - + \sqrt{2} } } } } \\ \: \: \: \: (k + 1) \: terms$

$\rm \: = \: \sqrt{2 + \underbrace{\sqrt{2 + \sqrt{2 + - - - + \sqrt{2} } } } } \\ \: \: \: \: k\: terms$

So, on substituting the value from step 2, we get

$\rm \: = \: \sqrt{2 + 2cos\bigg( \dfrac{\pi}{ {2}^{k + 1} } \bigg)}$

$\rm \: = \: \sqrt{2 \bigg[1+ cos\bigg( \dfrac{\pi}{ {2}^{k + 1} } \bigg)\bigg]}$

We know,

$\boxed{ \rm{ \:1 + cos2x = {2cos}^{2}x \: }}$

So, using this result, we get

$\rm \: = \: \sqrt{2 \times 2cos^{2} \bigg( \dfrac{1}{2} \times \dfrac{\pi}{ {2}^{k + 1} } \bigg)}$

$\rm \: = \: \sqrt{4cos^{2} \bigg( \dfrac{\pi}{ {2}^{k + 1 + 1} } \bigg)}$

$\rm \: = \: \sqrt{4cos^{2} \bigg( \dfrac{\pi}{ {2}^{k + 2} } \bigg)}$

$\rm \: = \: 2cos\bigg( \dfrac{\pi}{ {2}^{k + 2} } \bigg)$

$\rm\implies \:P(n) \: is \: true \: for \: n = k + 1$

Hence, By Principal of Mathematical Induction,

$\rm \: \sqrt{2 + \sqrt{2 + \sqrt{2 + - - - + \sqrt{2} } } } = 2cos\bigg( \dfrac{\pi}{ {2}^{n + 1} } \bigg)$

Answer 2

Answer:

$\sf \: \sqrt{2 + \sqrt{2 + \sqrt{2 + - - - + \sqrt{2} } } } = 2cos \bigg( \frac{\pi}{ {2}^{n + 1} } \bigg) \\ \\ \sf \: where \: root \: repeat \: n \: times \: and \: n \: is \: natural \: number$

$\\ \rule{300pt}{0.1pt}$

Step-by-step explanation:

$\\ \tt\operatorname{Let} P(n)=\sqrt{2+\sqrt{2+\sqrt{2+\ldots+\ldots+\sqrt{2}}}}=2 \cos \left(\frac{\pi}{2^{n+1}}\right) \qquad \ldots(i)$

For n=1 ,

$\text{L.H.S. of \( \tt (i)=\sqrt{2} \) and R.H.S. of \( \tt(i)=2 \cos \left(\dfrac{\pi}{2^{2}}\right) \)}$

$\\ \[ \begin{array}{l} \tt =2 \cos \left(\dfrac{\pi}{4}\right) \\ \\ \tt =2 \cdot \dfrac{1}{\sqrt{2}} \\ \\ =\sqrt{2} \end{array} \]$

Therefore, P(1) is true. Assume it is true for n=k ,

$\\ \tt P(k) \underset{(k \: \text{ radical \: sign})}{=\sqrt{2+\sqrt{2+\sqrt{2+\ldots+\ldots+\sqrt{2}}}}=2 \cos \left(\frac{\pi}{2^{k+1}}\right) }$

For n=k+1

$\begin{aligned} \tt P(k+1) & \tt \underset{ k + 1\text { radical sign }}{=\sqrt{2+\sqrt{2+\sqrt{2+\ldots+\ldots+\sqrt{2}}}}} \\ \\ & \tt=\sqrt{\{2+P(k)\}} \\ \\ & \tt=\sqrt{2+2 \cos \left(\frac{\pi}{2^{k+1}}\right)} \text { (By assumption step) } \\ \\ & \tt=\sqrt{2\left(1+\cos \left(\frac{\pi}{2^{k+1}}\right)\right)} \\ \\ & \tt=\sqrt{2\left(1+2 \cos ^{2}\left(\frac{\pi}{2^{k+1}}\right)-1\right)} \\ \\ & \tt=\sqrt{4 \cos ^{2}\left(\frac{\pi}{2^{k+2}}\right)} \\ \\ & \tt=2 \cos \left(\frac{\pi}{2^{k+2}}\right) \end{aligned}$

This shows that the result is true for n=k+1. Hence by the principle of mathematical induction, the result is true for all n ∈ N .