300 Followers Problem - Combinatorial Expressions!

It is well known that the generating functions for the sequences $\binom{2k}{k}$ and $(-1)^k \binom{2k}{k}$ for $k=0,1,2,\ldots$ are $\dfrac{1}{\sqrt{1-4x}}$ and $\dfrac{1}{\sqrt{1+4x}}$ respectively, i.e

$\dfrac{1}{\sqrt{1-4x}} = \displaystyle \sum_{k=0}^\infty \binom{2k}{k} x^k$ and $\dfrac{1}{\sqrt{1+4x}} = \displaystyle \sum_{k=0}^\infty (-1)^k \binom{2k}{k} x^k$

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$\dfrac{1}{\sqrt{1-4x}} \cdot \dfrac{1}{\sqrt{1-4x}} = \dfrac{1}{1-4x} = \displaystyle \sum_{k=0}^\infty (4x)^k$

We conclude that $A_n = 1 + 4 +4^2 + \ldots + 4^{2n} = \dfrac13 (4^{2n+1}-1)$

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Similarly, we can show for $B_n$ that $B_n = \displaystyle \sum_{k=0}^{2n} \left( \displaystyle \sum_{i+j=k} (-1)^k \binom{2i}{i} \binom{2j}{j} \right) \quad ... (2)$

and that the coefficients on the right hand side on (2) is precisely the coefficients of $x^0, x^1, x^2, \ldots , x^{2n}$ in the product of $\displaystyle \sum_{k=0}^\infty (-1)^k \binom{2k}{k}x^k$ .

Since $\dfrac{1}{\sqrt{1+4x}} \cdot \dfrac{1}{\sqrt{1+4x}} = \dfrac{1}{1+4x} = \displaystyle \sum_{k=0}^\infty (-1)^k (4x)^k$ , we conclude that $B_n = 1 - 4 + 4^2 - 4^3 + \ldots + 4^{2n} = \dfrac15 (4^{2n+1}+1)$

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Thus $5B_n - 3A_n = 2$ for all $n$ belonging to non-negative integers. Thus our answer is $\log_2(2) = \boxed{1.000}$ .