binomial inequality

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\hspace{-16}$Prove that\\\\ $\mathbf{\binom{2n+1}{1}.\binom{2n+1}{3}.\binom{2n+1}{5}.......\binom{2n+1}{2n-1}.\binom{2n+1}{2n+1}<\left(\frac{4^n-1}{n}\right)^n}

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Vivek @ Born this Way ·2011-11-28 00:23:08

We rewrite the above expression as for n>1

$\hspace{-16}$\\\\ $\mathbf{n\binom{2n+1}{1}.n\binom{2n+1}{3}.n\binom{2n+1}{5}...... .n\binom{2n+1}{2n-1}<\left(\frac{4^n-1}{1}\right)^n}$ .

From this, it is obvious that it is a matter of AM-GM.

ie.,

$\hspace{-16}$\\\\ $\mathbf{\sqrt[n]{n\binom{2n+1}{1}.n\binom{2n+1}{3}.n\binom{2n+1}{5}.......n\binom{2n+1}{2n-1}}<\left(\frac{4^n-1}{1}\right)}$ .

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Basic Integral Calculus Operators Symbols Matrix Cases

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∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

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° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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binomial inequality

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