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If $a+b=x$ and $ab=y$, express $a^n+b^n$ in terms of $x$ and $y$. The following may help you find the pattern. \begin{align*} a + b &= x\\ a^2 + b^2 &= x^2 - 2 y\\ a^3 + b^3 &= x^3 - 3 x y\\ a^4 + b^4 &= x^4 - 4 x^2 y + 2 y^2\\ a^5 + b^5 &= x^5 - 5 x^3 y + 5 x y^2\\ a^6 + b^6 &= x^6 - 6 x^4 y + 9 x^2 y^2 - 2 y^3\\ a^7 + b^7 &= x^7 - 7 x^5 y + 14 x^3 y^2 - 7 x y^3\\ a^8 + b^8 &= x^8 - 8 x^6 y + 20 x^4 y^2 - 16 x^2 y^3 + 2 y^4\\ a^9 + b^9 &= x^9 - 9 x^7 y + 27 x^5 y^2 - 30 x^3 y^3 + 9 x y^4\\ a^{10} + b^{10} &= x^{10} - 10 x^8 y + 35 x^6 y^2 - 50 x^4 y^3 + 25 x^2 y^4 - 2 y^5 \end{align*}

You might need more items to find the pattern. OK. The following code in Mathematica can save your time.

Table[{a^i + b^i, GroebnerBasis[{a^i + b^i, a + b - x, a b - y}, {x, y}, {a,b}]}, {i, 1, 20}] // TableForm
kiss my armpit
  • 3,896

2 Answers2

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  1. $a$ and $b$ are roots of the equation $T^2-xT+y=0$ and so you can use Newton's identities. This will give a recursive formula.

  2. A simpler recursion may be found as follows: Let $p_n=a^n+b^n$. Then $p_n x = (a^n+b^n)(a+b)=p_{n+1}+abp_{n-1}=p_{n+1}+yp_{n-1}$ and so $p_{n+1}=xp_n-yp_{n-1}$. The base cases are $p_0=2$ and $p_1=x$.

  3. OEIS can help you find a formula for the coefficients:

    http://oeis.org/search?q=2%2C5%2C9%2C14%2C20

    http://oeis.org/search?q=2%2C7%2C16%2C30%2C50

    They seem to be easily expressed as sums of two binomial coefficients. I've guessed that the $k$-column is given by $$ \binom{n+1-k}{n-2(k-1)}+\binom{n-k}{n-2(k-1)} $$

    You can also write the coefficients as a kind of Pascal triangle and guess the rule, which seems to be a variation of the one in the Pascal triangle. I've found this rule: $$ \binom{n+1}{k}'=\binom{n}{k}'+\binom{n}{k-1}'-\binom{n-2}{k-2}' $$

lhf
  • 216,483
0

$ a + b = x $

$ ab = y $

$ a-b = \sqrt {(a+b)^2 – 4ab} = \sqrt {x^2-4y} $

$ a= ({x + \sqrt {x^2-4y} })/2 $

$ b= (x - \sqrt {x^2-4y})/2 $

$ a^n + b^n $ = $ \{ (x + \sqrt {x^2-4y})^n $ + $ (x - \sqrt{x^2-4y}) ^n $ } / $2^n$

mahavir
  • 753
  • ... and when you expand those powers by the binomial theorem, the odd powers of $x^2-4y$ all cancel, leaving only even powers. – GEdgar Oct 04 '13 at 15:18