JEE Advanced 2023 Paper 1 · Mathematics · Sets, Relations and Functions · hard

A sequence is defined by a1=1 and an+1=an+1/an for n≥1. Find lfloor a100rfloor, where lfloor·rfloor denotes the greatest integer function.

Numerical answer

A sequence is defined by

a_1=1

and

a_{n+1}=a_n+\dfrac1{a_n}

for

n\ge1

. Find

\lfloor a_{100}\rfloor

, where

\lfloor\cdot\rfloor

denotes the greatest integer function.

Show answer and solution

Answer: 14

Square the recurrence:

a_{n+1}^2=a_n^2+2+\dfrac1{a_n^2}

. Summing from

n=1

99

a_{100}^2=1+2\cdot99+\sum_{n=1}^{99}\frac1{a_n^2}=199+\sum_{n=1}^{99}\frac1{a_n^2}.

a_{100}^2>199>196=14^2

. Also

a_n^2\ge 2n-1

by induction, so the sum is at most

1+\sum_{n=2}^{99}\frac1{2n-1}<1+\frac12\left(1+\ln 99\right)<4

(a crude bound), giving

a_{100}^2<203<225=15^2

.
Hence

14<a_{100}<15

(direct iteration gives

a_{100}\approx14.21

), and

\lfloor a_{100}\rfloor=14

. The key idea is that squaring turns a nonlinear recurrence into an almost arithmetic one, with a small correction term that is easy to bound from both sides.

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