Solve this:

Q2. The number of natural numbers n $\le$ 30 for which $\sqrt{\mathrm{n}+\sqrt{\mathrm{n}+\sqrt{\mathrm{n}+ ...............}}}$ is natural number is 
      (a) 30                       (b) zero                 (c) 6                          (d) 5 


 

Dear Student,

Please find below the solution to the asked query:

We know " Natural numbers = 1 , 2 , 3 , 4 , . . . " ( No negative numbers and no fractions. )

We assume :  x  = $\sqrt{n + \sqrt{n + \sqrt{n + . . . }}}$   --- ( 1 ) , Here " x "  is a natural number and " n $\le$ 30 " .

So,

$$\begin{gathered} x = \sqrt{n +\sqrt{n + \sqrt{n + \sqrt{n + . . . }}}} \\ \Rightarrow x = \sqrt{n +x} ( \mathrm{From} \mathrm{equation} 1 ) \\ \mathrm{Taking} \mathrm{whole} \mathrm{square} \mathrm{on} \mathrm{both} \mathrm{hand} \mathrm{side} \mathrm{and} \mathrm{get} : \\ \Rightarrow x^2 = {\left(\sqrt{n +x}\right)}^2 \\ \Rightarrow x^2 =n + x \\ \mathbf{\Rightarrow }{\mathit{x}}^{\mathbf{2}}\mathbf{ }\mathbf{-}\mathbf{ }\mathit{x}\mathbf{ }\mathbf{-}\mathbf{ }\mathit{n}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{2}\mathbf{ }\mathbf{)} \end{gathered}$$

From above equation we can see that we get value of ' x ' as natural number  when " n $\le$ 30 " and value of ' x ' is we get by using splitting the middle term method . We assume values of " n " as ( product of two numbers without considering their sign is equal to ' n ' ) and ( Sum of two numbers with considering their sign is equal to ' - 1 ' that is coefficient of ' x ' from equation 2 )

Then we get different cases to satisfy above conditions , As :

Case I : When " n = 2 " , So from equation 2 we get :

x² - x  - 2 = 0

$\Rightarrow$x² - 2 x + x - 2 = 0

$\Rightarrow$x ( x - 2 ) + 1 ( x - 2 ) = 0 

$\Rightarrow$( x - 2 )( x + 1 ) = 0  

So, x  =  2 and  - 1  , Here  ' 2 '  is a natural number so at " n = 2 " we get natural number from equation 1 .

Similarly :

Case II : When " n = 6 " , So from equation 2 we get :

x² - x  - 6 = 0  , We get x  = 3  and - 2 , Here  ' 3 '  is a natural number so at " n = 6 " we get natural number from equation 1 .

Case III : When " n = 12 " , So from equation 2 we get :

x² - x  - 12 = 0  , We get x  = 4  and - 3 , Here  ' 4 '  is a natural number so at " n = 12 " we get natural number from equation 1 .

Case IV : When " n = 20 " , So from equation 2 we get :

x² - x  - 20 = 0  , We get x  = 5  and - 4, Here  ' 5 '  is a natural number so at " n = 20 " we get natural number from equation 1 .

Case V : When " n = 30 " , So from equation 2 we get :

x² - x  - 30 = 0  , We get x  = 6  and - 5 , Here  ' 6 '  is a natural number so at " n = 30 " we get natural number from equation 1 .

Therefore,

Total number of natural number ' n ' can gives a natural number  = 5 .

So,

Option ( D )                                                                        ( Ans )


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