# If    $a_{n}=\sqrt{7+\sqrt{7+\sqrt{7+......}}}$    having $n$  radical signs then by methods of mathematical induction which is true Option 1) $a_{n}> 7,\forall \; n\geq 1\;$ Option 2) $\; \; a_{n}> 3,\forall \; n\geq 1\;$ Option 3) $\; \; a_{n}< 4,\forall \; n\geq 1\;$ Option 4) $\; \; a_{n}< 3,\forall \; n\geq 1$

As we learnt in

Steps of Mathematical Induction (Verification step) -

Step 1: Verification step

Actual verification of the proposition of the starting value $n=1$

- wherein

$2^{3n}-1$ is divisible by 7

Put n=1, It Satisfies.

and

Steps of Mathematical Induction (Induction Step) -

Step 2: Induction Step

Assuming the proposition to be true for n=k, and proving it is true for value     $n=k+1$

-

And

Steps of Mathematical Induction (Generalization Step) -

Combine Verification step and Induction step

$p(1)$ is true and $p(n)$ is true for $n+1$ assuming it is true for $n$

-

$a_{n} = \sqrt{7 +\sqrt{7 +\sqrt{7 +.....}}}$

$for\ n =1$

$a_{1} = \sqrt{7}$ which is less than 3 and lesser than 4

For n = k

ak < 7

$a_{k+1} = \sqrt{7 + a_{k}}$

$a_{k+1}^{2} = 7 + a_{k} < 7 +7 < 14$

$a_{k +1} < \sqrt{}14$

S= an < 7

Option 1)

$a_{n}> 7,\forall \; n\geq 1\;$

This option is incorrect.

Option 2)

$\; \; a_{n}> 3,\forall \; n\geq 1\;$

This option is correct.

Option 3)

$\; \; a_{n}< 4,\forall \; n\geq 1\;$

This option is incorrect.

Option 4)

$\; \; a_{n}< 3,\forall \; n\geq 1$

This option is incorrect.

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