Leysa {l}{frac{a_{m+1}}{2^m+1}+frac{a_{n-1}}{2^n-1}}{frac{a_{m+1}+4a_{m-1}}{2^n+1}}

\left. \begin{array} { l } { \frac { a _ { m + 1 } } { 2 ^ { m + 1 } } + \frac { a _ { n - 1 } } { 2 ^ { n - 1 } } } \\ { \frac { a _ { m + 1 } + 4 a _ { m - 1 } } { 2 ^ { n + 1 } } } \end{array} \right.

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First swap the first and second row. Then by subtracting the first and the second row from the third one and then subtracting a times the first row from the second we get: \left(\begin{array}{ccc|c} 1& a& 3 & b\\ 0 & 4-a^2 & -2a & -ab \\ 0 & 0 & -b^2-3 & -2 \end{array}\right) ...

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You may find the following easier to understand. Let |a| represents the order of a and let the g.c.d. of |a| and k be g. (a^k)^\frac{|a|}{g}=(a^{|a|})^\frac{k}{g}=e and so |a^k| ...

If a_n = \frac{{(n + 1)}^n}{n^{n - 1}} and c > 0 then \lim_{n \to \infty} n^{1-c}(a^c_n-a^c_{n-1}) =ce^c

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I think the approach in my answer to the question which you have linked should work without much change. To simplify typing I will use A = a_{n}, B = a_{n - 1}. Note that A/n, B/n \to e. I first ...

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I didn't check the detail of calculation but, if it is correct, the solution you have found is valid for a\neq 2 (i.e. the system has exactly a solution \forall a\neq2). Now you need also to ...

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