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The solution of $\sqrt{a_n} – 2\sqrt{a_{n-1}} + \sqrt{a_{n-2}} = 0$ where $a_0 = 1$ and $a_1 = 2$ is

  1. ${\Big[\frac{2^{n+1} + (-1)^n}{3}\Big]}^2$
  2. $(n+1)^2$
  3. $(n-1)^3$
  4. $(n-1)^2$
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Here, $a_0=1$, $a_1=2$ and

$\begin{align} \sqrt{a_r}-2\sqrt{a_{r-1}}+\sqrt{a_{r-2}}&=0;~[\text{Here }r \text{ is the simple variable replaced by }n]\\\Rightarrow \sqrt{a_r}-\sqrt{a_{r-1}}&=\sqrt{a_{r-1}}-\sqrt{a_{r-2}}\\\Rightarrow \sum_{r=2}^{n}(\sqrt{a_r}-\sqrt{a_{r-1}})&=\sum_{r=2}^{n}(\sqrt{a_{r-1}}-\sqrt{a_{r-2}});~[\text{Taking sum operator to both sides}] \\\Rightarrow \sqrt{a_n}-\sqrt{a_1}&=\sqrt{a_{n-1}}-\sqrt{a_0};~[\text{Telescoping sum, see link below in the note}]\\\Rightarrow \sqrt{a_n}-\sqrt{a_{n-1}}&=\sqrt{a_1}-\sqrt{a_0}=\sqrt{2}-1;~[\because a_1=2, a_0=1]\\\Rightarrow \sum_{r=1}^{n}(\sqrt{a_r}-\sqrt{a_{r-1}})&=\sum_{r=1}^{n}(\sqrt{2}-1);~[\text{Taking sum operator to both sides}]\\\Rightarrow \sqrt{a_n}-\sqrt{a_0}&=(\sqrt{2}-1)n\\\Rightarrow \sqrt{a_n}&=(\sqrt{2}-1)n+1;~[\because a_0=1\Rightarrow \sqrt{a_0}=1]\\ \therefore a_n&=\left\{ 1+(\sqrt{2}-1)n \right\}^2\end{align}$

 

The answer is not given in the options.

 

Note: To learn about Telescoping Sum, read it.

 

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