binomial sum ratio

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thewayofthe_dragon

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thewayofthe_dragon

#1 h Jun 16, 2024, 6:39 PM

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Someone please evaluate this ratio inside the log for any given n(I feel the sum doesn't have any nice closed form).

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thewayofthe_dragon

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thewayofthe_dragon

#2 h Jan 25, 2025, 2:29 PM

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Can someone solve?

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#3 h Apr 22, 2025, 1:08 PM

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thewayofthe_dragon wrote:

Can someone solve?

The argument of the logarithm equals to 1
So, option C

This post has been edited 1 time. Last edited by P162008, Apr 22, 2025, 1:09 PM
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#4 h Apr 23, 2025, 11:48 PM

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thewayofthe_dragon wrote:

Can someone solve?

Let $\zeta = \sum_{i=0}^{r} \binom{n}{2i}\binom{n-2i}{r-i}$

Now, define a generating function $\alpha(x)$ as $\alpha(x) = \sum_{r=0}^{\infty} \zeta x^r = \sum_{r=0}^{\infty} \sum_{i=0}^{r} \binom{n}{2i} \binom{n-2i}{r-i} x^r$

On swapping the order of summation, we get

$\alpha(x) = \sum_{i=0}^{\infty} \sum_{r=i}^{\infty} \binom{n}{2i} \binom{n-2i}{r-i} x^r = \sum_{i=0}^{\infty} \binom{n}{2i} x^i \sum_{r=i}^{\infty} \binom{n-2i}{r-i} x^{r-i}$

Let $r - i = t$ then $0 \leqslant r - i < \infty$ as $i \leqslant r < \infty$

Now, $\alpha(x) = \sum_{i=0}^{\infty} \binom{n}{2i} x^i \sum_{t=0}^{\infty} \binom{n-2i}{t} x^t = \sum_{i=0}^{\infty} \binom{n}{2i} x^i (1 + x)^{n - 2i} = (1 + x)^n \sum_{i=0}^{\infty} \binom{n}{2i} \left(\frac{\sqrt{x}}{1 + x}\right)^{2i}$

Then, $\alpha(x) = (1 + x)^n \left[\frac{(1 + \frac{\sqrt{x}}{1 + x})^n + (1 - \frac{\sqrt{x}}{1 + x})^n}{2}\right] = \frac{1}{2} \left[(1 + \sqrt{x} + x)^n + (1 - \sqrt{x} + x)^n\right]$

$\boxed{\therefore \zeta = \left[x^r\right] \frac{1}{2} \left[(1 + \sqrt{x} + x)^n + (1 - \sqrt{x} + x)^n\right]}$

Similarly, Let $\xi = \sum_{i=r}^{n} \binom{n}{i} \binom{2i}{2r} \left(\frac{1}{2}\right)^{2i - 2r} \left(\frac{3}{4}\right)^{n- i}$

Again, define a generating function $\beta(x)$ as $\beta(x) = \sum_{r=0}^{\infty} \xi x^r = \sum_{r=0}^{\infty} \sum_{i=r}^{n} \binom{n}{i} \binom{2i}{2r} \left(\frac{1}{2}\right)^{2i-2r} \left(\frac{3}{4}\right)^{n- i} x^r$

On swapping the order of summation, we get

$\beta(x) = \sum_{i=0}^{\infty} \sum_{r=0}^{i} \binom{n}{i} \binom{2i}{2r} \left(\frac{1}{2}\right)^{2i-2r} \left(\frac{3}{4}\right)^{n- i} x^r = \sum_{i=0}^{\infty} \binom{n}{i} \left(\frac{3}{4}\right)^{n - i} \left(\frac{1}{4}\right)^i \sum_{r=0}^{i} \binom{2i}{2r} \left(2\sqrt{x}\right)^{2r}$

$\beta(x) = \sum_{i=0}^{\infty} \binom{n}{i} \left(\frac{3}{4}\right)^{n - i} \left(\frac{1}{4}\right)^i \left[\frac{(1 + 2\sqrt{x})^{2i} + (1 - 2\sqrt{x})^{2i}}{2}\right] = \sum_{i=0}^{\infty} \binom{n}{i} \left(\frac{3}{4}\right)^{n - i} \left(\frac{1}{4}\right)^i \left[\frac{(4x + 4\sqrt{x} + 1)^i + (4x - 4\sqrt{x} + 1)^i}{2}\right]$

Then, $\beta(x) = \frac{1}{2} \left[\sum_{i=0}^{\infty} \binom{n}{i} \left(\frac{3}{4}\right)^{n-i} \left(x + \sqrt{x} + \frac{1}{4}\right)^i + \sum_{i=0}^{\infty} \binom{n}{i} \left(\frac{3}{4}\right)^{n-i} \left(x - \sqrt{x} + \frac{1}{4}\right)^i \right] = \frac{1}{2} \left[(1 + \sqrt{x} + x)^n + (1 - \sqrt{x} + x)^n\right]$

$\boxed{\therefore \xi = \left[x^r\right] \frac{1}{2} \left[(1 + \sqrt{x} + x)^n + (1 - \sqrt{x} + x)^n\right]}$

Finally, $\boxed{\log_{10} \left(\frac{\zeta}{\xi}\right) = \log_{10} 1 = \boxed{0}}$ :love:

:love:

This post has been edited 18 times. Last edited by P162008, Apr 28, 2025, 2:08 PM
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