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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.

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# Recursive series for sum of 2 powers

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