21Verify that \sqrt{4n+1} \le \sqrt{n} + \sqrt{n+1} \le \sqrt{4n+2}
Note that for any natural number N, we have \left \lfloor N+1 \right \rfloor > \left \lfloor N \right \rfloor iff N+1 is a perfect square.
But since No square is congruent to 2 or 3 modulo 4 , we always have \left \lfloor 4n+1 \right \rfloor = \left \lfloor 4n+2 \right \rfloor = \left \lfloor 4n+3 \right \rfloor
So by the non decreasing nature of floor, \left \lfloor 4n+1 \right \rfloor = \left \lfloor 4n+2 \right \rfloor = \left \lfloor 4n+3 \right \rfloor =\left \lfloor \sqrt{n} + \sqrt{n+1} \right \rfloor
11Nice solution Shubhodip [1]
Lets take a2≤n<(a+1)2
Using the fact that \left[a+\sqrt{a^2+1} \right]=2a we have
\left[\sqrt{n}+\sqrt{n+1} \right]=2a \text{ or } 2a+1.... (*)
Similarly we also have \left[\sqrt{4n+1} \right]=2a \text{ or } 2a+1.... (**)
Now let n=a2+b
So \left[\sqrt{n}+\sqrt{n+1} \right]=2a+1 \text{ if } b\ge a+1
and 2a if b≤a-1
Now b\ge a+1 \Leftrightarrow 4n+1\ge (2a+1)^2\Rightarrow \left[\sqrt{4n+1} \right]=2a+1
Similarly b\le a-1 \Leftrightarrow 4n+1\le 4a^2+4a-4<4a^2+4a+1=(2a+1)^2\Rightarrow \left[\sqrt{4n+1} \right]=2a
Now when b=a, 4n+1=(2a+1)2
\left[\sqrt{n}+\sqrt{n+1} \right]=\left[\sqrt{a^2+a}+\sqrt{a^2+a+1} \right]>\left[a+a \right] implying
\left[\sqrt{n}+\sqrt{n+1} \right]=2a+1
21
Shubhodip
·2012-01-20 08:46:23
sorry
i have forgot to add square roots in most of the places where it was required
so add them when you feel things are wrong :)
