Question

If $a_n=\sqrt{7+\sqrt{7+\sqrt{7+\cdots }}}n$ times using the principle of mathematical induction which of the following is true ?

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

Answer 1

The correct option is C $a_n<4,\forall n\ge 1$

$p\left(n\right):a_n=\sqrt{7+\sqrt{7+\sqrt{7+\cdots }}}n$times
For $n=1,a_1=\sqrt{7}.$ So, $a_n>7$ is false.
$a_1>3$ is false. $a_1=\sqrt{7}<4$ and $a_1=\sqrt{7}<3$ is true.
$a_2=\sqrt{7+\sqrt{7}}$
Here, $2<\sqrt{7}<3\Rightarrow 9<7+\sqrt{7}<10<16\Rightarrow 3<\sqrt{7+\sqrt{7}}<4\Rightarrow 3<a_2<4$
$\therefore a_2<3$ is false. But, $a_2<4$ is true.
Thus, $a_1<4,a_2<4$ is true.
Suppose $a_k<4$
Now $a_{k+1}=\sqrt{7+\sqrt{7+\sqrt{7+\cdots }}}(k+1)$times
$a_{k+1}=\sqrt{7+a_k}$
$\therefore a_{k+1}^2=7+a_k<7+4,\text{So}{a}_{k+1}<\sqrt{11}<4$So,$a_{k+1}<4$
Thus $a_n<4$ is true for $n=k+1$
$\therefore a_n<4,\forall n\in N$