[JEE Advanced 1987] Prove by induction that $\dfrac{(2n)!}{2^{2n}(n!)^2}\le\dfrac1{\sqrt{3n+1}}$ for all positive integers $n$.
Prove by induction that $\dfrac{(2n)!}{2^{2n}(n!)^2}\le\dfrac1{\sqrt{3n+1}}$ for all positive integers $n$.
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VAVidaara Admin
✓ Vidaara Team
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· 1mo ago
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Answer: Proved.
Base $n=1$: $\frac12\le\frac12$. The step requires $\frac{2n+1}{2n+2}\le\sqrt{\frac{3n+1}{3n+4}}$, which holds since cross-multiplying gives $(2n+1)^2(3n+4)\le(2n+2)^2(3n+1)$.
JEE Advanced 1987 · Binomial Theorem — verified solution by the Vidaara Team.
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