Has anyone seen this before? It must have been noted before!(adsbygoogle = window.adsbygoogle || []).push({});

Let c = a+b and d = ab Then the series [tex] S_{n} = c*S_{n-1} - d*S_{n-2}[/tex] having the values 2 and c for [tex]S_0[/tex] and [tex]S_1[/tex] has the explicit formula

[tex]S_{n} = a^{n} + b^{n}[/tex].

Proof

The Explicit formula for the above recursive series is found by finding the roots e and f of the equation [tex]u^{2} - c*u + d = 0[/tex] and solving the set of equations:

[tex] 2 = A + B[/tex] and

[tex] c = Ae + Bf[/tex] from which we determine that

[tex] S_{n} = Ae^{n} + Bf^{n}[/tex]

We need to show that A = B = 1, e = a and f = b.

Now e,f = (c +/- Sq root[(c)^{2} - 4ab])/2 = (a+b +/- (a-b))/2 = a,b

Solving the set of equations we also see that A=B=1. Q.E.D.

Most likely the proof could be done by induction also, but I wanted to focus on the general theory of recursive series.

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Recursive series for sum of 2 powers

Loading...

Similar Threads - Recursive series powers | Date |
---|---|

Converting an explicit series to a recursive form | Aug 12, 2014 |

All Square Numbers Follow a Recursive Series? | Jun 19, 2011 |

Recursive series S(n): 8*N*S(n) +1 = square | Nov 18, 2010 |

New recursive series identity | Oct 17, 2007 |

Recursive Series Equality | Oct 4, 2007 |

**Physics Forums - The Fusion of Science and Community**