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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. $(n-1)^2$
in Combinatory by Boss (12.9k points)
edited by | 129 views
0
Ans $a_{n}=\sqrt{c_{1}+c_{2}n}$

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

then equation will be $A_{n}-2.A_{n-1}+A_{n-2}=0$
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 $b_{1} = A + B \Rightarrow B = \sqrt {2} - 1 \ (\because b_{1} = \sqrt a_{1} = \sqrt 2 )$

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

$\therefore a_{n} = [1 + (\sqrt 2 -1)n]^2$
0
You're somewhat wrong. Here, in your answers A and B values are 1.

If you'll calculate, then you'll get b = n+1.

Then b= $\sqrt{a}$

Put value of b in the above equation.

You'll get a=(n+1)²

1 Answer

+1 vote

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.

 

by Active (3.1k points)
+1
Can you check $\mathrm A$ option by substituting the values?
0

For the option A, if we take $a_n=\left[ \frac{2^{n+1}+(-1)^n}{3} \right]^2$, then

$a_1=\left[\frac{2^2+(-1)^1}{3}\right]^2=\left[\frac{3}{3}\right]^2=1$

But it's given that $a_1=2$. Therefore, the option A must be wrong.

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