take $a_{n}=A_{n}^{2}$

then equation will be $A_{n}-2.A_{n-1}+A_{n-2}=0$

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+1 vote

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

- ${\Big[\frac{2^{n+1} + (-1)^n}{3}\Big]}^2$
- $(n+1)^2$
- $(n-1)^3$
- $(n-1)^2$

closed with the note:
Solved.

0

Ans $a_{n}=\sqrt{c_{1}+c_{2}n}$

take $a_{n}=A_{n}^{2}$

then equation will be $A_{n}-2.A_{n-1}+A_{n-2}=0$

take $a_{n}=A_{n}^{2}$

then equation will be $A_{n}-2.A_{n-1}+A_{n-2}=0$

0

I think we are not allowed to post coaching institute questions.

https://gateoverflow.in/blog/8120/questions-from-coaching-institutes

0

Let, $b_{n} = \sqrt a_{n}$

$\therefore b_{n-1} = \sqrt a_{n-1}$

$\therefore b_{n-2} = \sqrt a_{n-2}$

$\therefore \sqrt a_{n} -2 \sqrt a_{n-1} + \sqrt a_{n-2} = 0 \Rightarrow b_{n} -2b_{n-1} + b_{n-2} = 0$

Now, its characteristic equation is $x^{2} -2x +1 = 0 \Rightarrow x = 1, 1$

$\therefore$ General solution is $b_{n} = (A + Bn)x^{n} = (A + Bn)1^{n} = A + Bn$ where $A, B$ are arbitrary constants.

Putting $n=0$ and $n=1$, we get $b_{0} = A \Rightarrow A = 1 \ (\because b_{0} = \sqrt a_{0} = 1)$

Putting $n=0$ and $n=1$, we get $b_{1} = A + B \Rightarrow B = \sqrt {2} - 1 \ (\because b_{1} = \sqrt a_{1} = \sqrt 2 )$

$\therefore b_{n} = 1 + (\sqrt 2 -1)n$

$\therefore a_{n} = [1 + (\sqrt 2 -1)n]^2$

$\therefore b_{n-1} = \sqrt a_{n-1}$

$\therefore b_{n-2} = \sqrt a_{n-2}$

$\therefore \sqrt a_{n} -2 \sqrt a_{n-1} + \sqrt a_{n-2} = 0 \Rightarrow b_{n} -2b_{n-1} + b_{n-2} = 0$

Now, its characteristic equation is $x^{2} -2x +1 = 0 \Rightarrow x = 1, 1$

$\therefore$ General solution is $b_{n} = (A + Bn)x^{n} = (A + Bn)1^{n} = A + Bn$ where $A, B$ are arbitrary constants.

Putting $n=0$ and $n=1$, we get $b_{0} = A \Rightarrow A = 1 \ (\because b_{0} = \sqrt a_{0} = 1)$

Putting $n=0$ and $n=1$, we get $b_{1} = A + B \Rightarrow B = \sqrt {2} - 1 \ (\because b_{1} = \sqrt a_{1} = \sqrt 2 )$

$\therefore b_{n} = 1 + (\sqrt 2 -1)n$

$\therefore a_{n} = [1 + (\sqrt 2 -1)n]^2$

+2 votes

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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