[JEE Advanced 2000] prove that for every positive integer n sqrt 4n 1 sqrt n sqrt n
Prove that for every positive integer $n$, $\sqrt{4n+1}<\sqrt n+\sqrt{n+1}<\sqrt{4n+2}$, and hence $[\sqrt n+\sqrt{n+1}]=[\sqrt{4n+1}]$.
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· 4d ago
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Answer: Proved.
Squaring, $(\sqrt n+\sqrt{n+1})^2=2n+1+2\sqrt{n(n+1)}$; since $4n+1<2n+1+2\sqrt{n(n+1)}<4n+2$, the integer parts of $\sqrt n+\sqrt{n+1}$ and $\sqrt{4n+1}$ coincide (no perfect square lies between).
JEE Advanced 2000 · Binomial Theorem — verified solution by the Vidaara Team.
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