Question

If ${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .....} } } $ having $n$ radical signs, then by method of mathematical indication which of the following is true?
A.\[{a_n} > 6,\forall n > 1\]
B.\[{a_n} > 3,\forall n > 1\]
C.\[{a_n} > 4,\forall n > 1\]
D.\[{a_n} > 2,\forall n > 1\]

Answer 1

Hint: We have given that ${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + } ....} } $. We have to check which of the conditions satisfy mathematical induction. For this firstly we have to take the repeating term here repeating term is $\sqrt 7 $. We put it equal to ${a_n}$. So we get ${a_n} = \sqrt {7 + {a_n}} $ . We put it equal to ${a_n}$. So we get ${a_n} = \sqrt {7 + {a_n}} $. We convert it into a quadratic equation. Now we solve this quadratic equation by quadratic formula method. Since quadratic equation has degree two there for its solution will give us two roots. We check the roots with the given option if any option matches it will be our solution.

Complete step-by-step answer:
We have given that ${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + } ....} } $ having $n$ radical signs
Since $\sqrt 7 $ is repeating is self there ${a_n} = \sqrt {7 + {a_n}} $
Squaring both sides, we get
${a^2}_n = 7 + {a_n}$
${a_n}^2 = {a_n} - 7 = 0$ ……………….. \[\left( i \right)\]
Now this is a quadratic equation with variable${a_n}$.
Now we will find the value of \[{a_n}\]by quadratic formula method.
Comparing \[\left( i \right)\]with \[ax_n^{^2} = b{x_n} + C = 0\]
$a = 1,b = - 1,c = - 7$
The quadratic formula is
${a_n} = \dfrac{{b + \sqrt {{b^2} - 4ac} }}{{2a}}$
$\Rightarrow$ \[{a_n} = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4 \times ( - 1)} }}{{2 \times 1}}\]
$\Rightarrow$ \[{a_n} = \dfrac{{1 \pm \sqrt {1 + 28} }}{2}\]
$\Rightarrow$ \[{a_n} = \dfrac{{1 \pm \sqrt {29} }}{2}\]
So we have two values of ${a_n}$ that are
\[{a_n} = \dfrac{{ + 1 + \sqrt {29} }}{2}\] and \[{a_n} = \dfrac{{ + 1 - \sqrt {29} }}{2}\]
$\Rightarrow$ \[{a_n} = \dfrac{{ + 1 + 5.39}}{2}\] and \[{a_n} = \dfrac{{ + 1 - 5.39}}{2}\]
$\Rightarrow$ \[{a_n} = \dfrac{{6.39}}{2}\] and \[{a_n} = \dfrac{{ - 4.39}}{2}\]
\[{a_n} = 3.195\] and \[{a_n} = - 2.195\]
Now we have ${a_n} > 0$ therefore we neglect \[{a_n} = - 2.195\]
Therefore \[{a_n} = 3.195\]
Which means ${a_n} > 3$ $\sqrt n > 1$
Option $(B)$ is correct.

Note: Mathematical induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number. The technique involves two steps to prove the statement.
Step$1$– It proves that the statement is true for initial value.
Step$2$– It proves that if the statement is true for the \[{n^{th}}\]iteration then it will also true for \[{(n + 1)^{th}}\]iteration