IIT-JEE 2000Question 1
For every positive integer $n$, prove that $$\sqrt{4n+1}<\sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}.$$ Hence or otherwise, prove that $$\left[\sqrt{n}+\sqrt{n+1}\right]=\left[\sqrt{4n+1}\right],$$ where $[x]$ denotes the greatest integer not exceeding $x$.
Sequence and Series · Miscellaneous
Answerproved
Solution
Formulas used. (1) $n^2<n^2+n<\left(n+\frac12\right)^2$(2) Every perfect square leaves remainder $0$ or $1$ on division by $4$Step 1 - Square the middle. $$\left(\sqrt n+\sqrt{n+1}\right)^2=2n+1+2\sqrt{n^2+n}.$$ Step 2 - Bound $\sqrt{n^2+n}$. By formula (1), $n<\sqrt{n^2+n}<n+\frac12$. So $$4n+1<\left(\sqrt n+\sqrt{n+1}\right)^2<4n+2,$$ which gives $\sqrt{4n+1}<\sqrt n+\sqrt{n+1}<\sqrt{4n+2}$. Step 3 - Greatest integer. Suppose $\left[\sqrt{4n+2}\right]>\left[\sqrt{4n+1}\right]$. Then some integer $m$ has $\sqrt{4n+1}<m\le\sqrt{4n+2}$, i.e. $m^2=4n+2$. But $4n+2$ leaves remainder $2$ on division by $4$, and no square does. So $$\left[\sqrt{4n+1}\right]=\left[\sqrt{4n+2}\right].$$ Step 4. $\sqrt n+\sqrt{n+1}$ lies between these two numbers, which have the same integer part, so $$\left[\sqrt n+\sqrt{n+1}\right]=\left[\sqrt{4n+1}\right].$$ Answer: proved