If `a_n = sqrt(7+ sqrt( 7+sqrt7+ ....... )))` having n radical signs then by methods of mathematical induction which is true

Options

A. `a_n gt 7,forall n ge 1`
B. `n_ngt 3,foralln ge 1`
C. `a_nlt 4, forall n ge 1`
D. `a_nlt 3,forall n ge 1`

Explanation

Let `P(n):a_(n)=sqrt(7+sqrt(7+sqrt(7+....)))` (n radical sign)
Step I For `n=1`,
`P(1):a_1=sqrt(7)lt 4`
Step II Assume that `a_(k)lt 4` for all natural number ,`n=k`
Step III For `n=k+1`,
`P(k+1):a_(k+1)sqrt(7+sqrt(7+sqrt(7+....)))` [(k+1)radical sign]
`=sqrt(7+a_k)lt sqrt(7+4)[becuae a_klt 4]`
`lt 4` [by assumption]
This shows that `a_(k+1)lt 4`, i.e., the result is true for `n=k+1`. Hence , by the principle of mathematial induction
`a_nlt 4, forall n ge 1`