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If: \begin{eqnarray*} ax+by&=&m_1,\\ ax^2+by^2&=&m_2,\\ ax^3+by^3&=&m_3,\\ ax^4+by^4&=&m_4\\ &\vdots& \\ \end{eqnarray*} Where : $m_1,m_2,m_3,m_4$ are constants

What is the general formula for :$ax^n+by^n$

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What are you trying to do? This is not a very clear question. Unless you think $ax^n + by^n = m_n$ is a valid answer. – mrf Aug 31 '12 at 9:21
$a = b = m_k = 0$ – draks ... Aug 31 '12 at 9:26
Seems like a well-posed question. I assume you derive the unknowns a, b, x, y. – Gerenuk Aug 31 '12 at 11:16
I think $ax^n + by^n$ is a pretty nice form already. – Graphth Aug 31 '12 at 12:43

Here is a suggestion for a simplification. You could try writing $x=t+u$ and $y=t-u$. From that it follows that

$abt^2\sum_{k=2}^n \binom{n}{k}m_1^{n-k}t^{k-2}\left(b^{k-1}+(-1)^ka^{k-1}\right)=(a+b)^nm_n-m_1^n$

Basically rearrange the first equation for $t$ and plug in. Hope I didn't do a mistake :) This looks long but might make the algebra easier, however it's still cumbersome.

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The question is not clear, but maybe what's wanted is a formula for $m_n$ for all $n$ in terms of previous $m_k$. Note $$xym_1=ax^2y+bxy^2\tag1$$ and $$(x+y)m_2=ax^3+by^3+ax^2y+bxy^2=m_3+xym_1\tag2$$ so we have $$m_3=(x+y)m_2-xym_1\tag3$$ In fact, we have in general $$m_n=(x+y)m_{n-1}-xym_{n-2}\tag4$$ If this isn't what's wanted, please edit a clarification into the problem.

EDIT: Gerenuk's suggestion in the comments is very good, though it seems a bit messy to carry out. Maybe I'm not seeing the neatest way to arrange the calculations. One can rewrite (4) as $$x+y={m_n+xym_{n-2}\over m_{n-1}}\tag5$$ and also $$x+y={m_{n-1}+xym_{n-3}\over m_{n-2}}\tag6$$ then set $(5)=(6)$ and solve for $xy$: $$xy={m_nm_{n-2}-m_{n-1}^2\over m_{n-1}m_{n-3}-m_{n-2}^2}\tag7$$ Now put (7) into (5) to get an expression for $x+y$ in terms of the $m_i$, and then put that expression and (7) into (4) to get an equation just relating the $m_i$.

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That's really clever! Now rearrange two of these equations for x and y and you get your solution! :) – Gerenuk Aug 31 '12 at 14:20

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