Question:
If
\[
a_n=
\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}
\]
(\(n\) radicals), then which of the following is true?
If
\[
a_n=
\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}
\]
(\(n\) radicals), then which of the following is true?
Show Hint
For nested radical sequences, try finding a simple upper or lower bound. Mathematical induction is often the fastest method to prove inequalities involving recursively defined sequences.
Updated On: Jun 10, 2026
- \(a_n>7 \quad \forall n\ge1\)
- \(a_n>3 \quad \forall n\ge1\)
- \(a_n<3 \quad \forall n\ge1\)
- \(a_n<4 \quad \forall n\ge1\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
Nested radical sequences are often analyzed using mathematical induction and bounding techniques.
The sequence
\[
a_n=
\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}
\]
contains only positive quantities. We attempt to establish an upper bound.
Step 1: Verify the first term For \(n=1\), \[ a_1=\sqrt7 \] Since \[ \sqrt7\approx2.646 \] we obtain \[ a_1<4 \] Thus the statement is true for the first term.
Step 2: Assume \(a_n<4\) Suppose for some \(n\), \[ a_n<4 \] We now examine \(a_{n+1}\). \[ a_{n+1} = \sqrt{7+a_n} \] Using the assumption, \[ a_{n+1} < \sqrt{7+4} \] \[ a_{n+1} < \sqrt{11} \] Since \[ \sqrt{11}\approx3.316<4 \] therefore \[ a_{n+1}<4 \]
Step 3: Apply induction The statement is true for \(n=1\) and if true for \(n\), then true for \(n+1\). Hence by mathematical induction, \[ a_n<4 \] for every positive integer \(n\).
Step 4: Check other options Clearly, \[ a_1=\sqrt7<3 \] is false. Hence \(a_n>3\) for all \(n\) cannot be true. Similarly, \[ a_n>7 \] is impossible because every term is obtained by taking square roots. Therefore the only correct statement is \[ \boxed{a_n<4 \quad \forall n\ge1} \]
Step 1: Verify the first term For \(n=1\), \[ a_1=\sqrt7 \] Since \[ \sqrt7\approx2.646 \] we obtain \[ a_1<4 \] Thus the statement is true for the first term.
Step 2: Assume \(a_n<4\) Suppose for some \(n\), \[ a_n<4 \] We now examine \(a_{n+1}\). \[ a_{n+1} = \sqrt{7+a_n} \] Using the assumption, \[ a_{n+1} < \sqrt{7+4} \] \[ a_{n+1} < \sqrt{11} \] Since \[ \sqrt{11}\approx3.316<4 \] therefore \[ a_{n+1}<4 \]
Step 3: Apply induction The statement is true for \(n=1\) and if true for \(n\), then true for \(n+1\). Hence by mathematical induction, \[ a_n<4 \] for every positive integer \(n\).
Step 4: Check other options Clearly, \[ a_1=\sqrt7<3 \] is false. Hence \(a_n>3\) for all \(n\) cannot be true. Similarly, \[ a_n>7 \] is impossible because every term is obtained by taking square roots. Therefore the only correct statement is \[ \boxed{a_n<4 \quad \forall n\ge1} \]
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