Question

$\mathrm{If}{a}_n=\sqrt{7+\sqrt{7+\sqrt{7+......}}}$
having n radical signs, then by the principle of mathematical induction, which of the following option is true?

• A. $a_n>7\forall n\ge 1$
• B. $a_n<7\forall n\ge 1$
• C. $a_n<4\forall n\ge 1$
• D. $a_n<3\forall n\ge 1$

Answer 1

The correct option is B $a_n<7\forall n\ge 1$

$\mathrm{Given}{a}_n=\sqrt{7+\sqrt{7+\sqrt{7+......}}}\mathrm{Then}{a}_1=\sqrt{7}<7a_2=\sqrt{7+\sqrt{7}}<7\mathrm{Let}\mathrm{us}\mathrm{assume}{a}_n<7...\left(1\right)\mathrm{Then}{a}_{n+1}=\sqrt{7+a_n}\Rightarrow {a^2}_{n+1}=7+a_n\Rightarrow {a^2}_{n+1}<7+7=14\left(\mathrm{From}\left(1\right)\right]\Rightarrow a_{n+1}<\sqrt{14}<7\mathrm{So}{a}_{n+1}<7\mathrm{whenever}{a}_n<7\mathrm{Thus},\mathrm{by}\mathrm{the}\mathrm{principle}\mathrm{of}\mathrm{mathematical}\mathrm{induction},a_n<7,\forall n\in N$