Version 1. Let $n$ be a positive integer, and set $N=2^{n}$. Determine the smallest real number $a_{n}$ such that, for all real $x$, $$\sqrt[N]{\frac{x^{2N}+1}{2}}\leqslant a_n(x-1)^2+x.$$ Please provide the value of N when $a_n = N/2$.\n\nVersion 2. For every positive integer $N$, determine the smallest real number $b_{N}$ such that, for all real $x$, $$\sqrt[N]{\frac{x^{2N}+1}{2}}\leqslant b_N(x-1)^2+x.$$ Please provide the sum of  $a_{4}$$b_{6}$.

Let's think step by step and output the final answer within \boxed{}.
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