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Pranav
Well-known member
- Nov 4, 2013
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Problem:
Prove:
$$\sqrt{C_1}+\sqrt{C_2}+\sqrt{C_3}+......+\sqrt{C_n} \leq 2^{n-1}+\frac{n-1}{2}$$
where $C_0,C_1,C_2,....,C_n$ are combinatorial coefficients in the expansion of $(1+x)^n$, $n \in \mathbb{N}$.
Attempt:
I thought of using the RMS-AM inequality and got:
$$\sqrt{C_1}+\sqrt{C_2}+\sqrt{C_3}+.....+\sqrt{C_n} \leq \sqrt{\frac{2^n-1}{n}}$$
Where I have used $C_1+C_2+C_3+.....+C_n=2^n-1$.
But I don't see how to proceed from here.
Any help is appreciated. Thanks!
Prove:
$$\sqrt{C_1}+\sqrt{C_2}+\sqrt{C_3}+......+\sqrt{C_n} \leq 2^{n-1}+\frac{n-1}{2}$$
where $C_0,C_1,C_2,....,C_n$ are combinatorial coefficients in the expansion of $(1+x)^n$, $n \in \mathbb{N}$.
Attempt:
I thought of using the RMS-AM inequality and got:
$$\sqrt{C_1}+\sqrt{C_2}+\sqrt{C_3}+.....+\sqrt{C_n} \leq \sqrt{\frac{2^n-1}{n}}$$
Where I have used $C_1+C_2+C_3+.....+C_n=2^n-1$.
But I don't see how to proceed from here.
Any help is appreciated. Thanks!
Last edited: