Zgjidh {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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5 probleme të ngjashme me:

Probleme të ngjashme nga kërkimi në ueb

For which values of a,b\in\mathbb{R} does the matrix have no solutions, one solution and infinite solutions.

https://math.stackexchange.com/questions/2018447/for-which-values-of-a-b-in-mathbbr-does-the-matrix-have-no-solutions-one-so

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) ...

How to prove this equation with Bessel function and Laguerre function?

https://math.stackexchange.com/questions/2541322/how-to-prove-this-equation-with-bessel-function-and-laguerre-function

Coordinate matrix of a vector in terms of a basis

https://math.stackexchange.com/questions/108389/coordinate-matrix-of-a-vector-in-terms-of-a-basis

(I'll do it for dimension n because the difficulty is the same) Suppose you write \{e_1,\ldots,e_n\} for the canonical basis (could be any basis, actually), and let \{f_1,\ldots,f_n\} be another ...

How can 1=\frac{ord\: a}{gcd(ord\:a,k)}|\frac{k}{gcd(ord\:a,k)}?

https://math.stackexchange.com/questions/2530700/how-can-1-fracord-agcdord-a-k-frackgcdord-a-k

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

https://math.stackexchange.com/questions/2259452/if-a-n-fracn-1nnn-1-and-c-0-then-lim-n-to-infty

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 ...

Solve the system of linear equations for all values of a

https://math.stackexchange.com/q/2697250

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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