The sequence #0, 2, 8, 30, 112, 418,...# is defined recursively by #a_0 = 0#, #a_1 = 2#, #a_(n+1) = 4a_n-a_(n-1)#. What is the formula for a general term #a_n#?

1 Answer
George C.
Sep 16, 2016

#a_n =((2+sqrt(3))^n - (2-sqrt(3))^n)/sqrt(3)#

Explanation:

#color(white)()#
Proof by induction

Case #n=0#

#((2+sqrt(3))^0 - (2-sqrt(3))^0)/sqrt(3) = (1 - 1) / sqrt(3) = 0 = a_0#

Case #n=1#

#((2+sqrt(3))^1 - (2-sqrt(3))^1)/sqrt(3) = (2sqrt(3))/(sqrt(3)) = 2 = a_1#

Induction step

Note that #(2-sqrt(3))(2+sqrt(3)) = 2^2-(sqrt(3))^2 = 4-3 = 1#

So:

#((2+sqrt(3))^(n+1)-(2-sqrt(3))^(n+1))/sqrt(3)#

#=((2+sqrt(3))(2+sqrt(3))^n-(2-sqrt(3))(2-sqrt(3))^n)/sqrt(3)#

#=(4((2+sqrt(3))^n-(2-sqrt(3))^n)-((2-sqrt(3))(2+sqrt(3))^n-(2+sqrt(3))(2-sqrt(3))^n))/sqrt(3)#

#=(4((2+sqrt(3))^n-(2-sqrt(3))^n))/sqrt(3)-((2+sqrt(3))^(n-1)-(2-sqrt(3))^(n-1))/sqrt(3)#

#=4a_n-a_(n-1)#

#color(white)()#
Bonus

#3a_n^2+4# is always the square of an integer ...

Note that #(2+sqrt(3))^n(2-sqrt(3))^n = 1^n = 1#

So:

#3a_n^2+4 = 3(((2+sqrt(3))^n - (2-sqrt(3))^n)/sqrt(3))^2 + 4#

#color(white)(3a_n^2+4) = ((2+sqrt(3))^n - (2-sqrt(3))^n)^2 + 4#

#color(white)(3a_n^2+4) = (2+sqrt(3))^(2n) - 2(2+sqrt(3))^n(2-sqrt(3))^n+(2-sqrt(3))^(2n) + 4#

#color(white)(3a_n^2+4) = (2+sqrt(3))^(2n) + 2(2+sqrt(3))^n(2-sqrt(3))^n+(2-sqrt(3))^(2n)#

#color(white)(3a_n^2+4) = ((2+sqrt(3))^n + (2-sqrt(3))^n)^2#

Note that #(2+sqrt(3))^n+(2-sqrt(3))^n# must be an integer, since odd powers of #sqrt(3)# from the two binomial powers carry opposite signs.

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