# Generating function of a recurrance relation

## Main Question or Discussion Point

Suppose A(x) is a generating function for the sequence a0, a1, a2, . . . that satisfies
the recurrence a[n+2] = −a[n+1] + 6a[n] for n > 0, with initial conditions a = 2 and
a = −1. Find a formula for A(x) and use it to find an explicit formula for a[n].

I dont know what im doing wrong, here is what I have done..
the first few terms of this sequence are 2,-1,13,-19,97,-211
my taylor series expansion is .. 1 + x + 5 x^2 + x^3 + 29x^4 which is wrong..

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a[n+2] = −a[n+1] + 6a[n]

all summations for n_> 0

∑ a[n+2] = -∑a[n+1] + 6∑a[n]

∑ a[n+2]x^n = -∑a[n+1]x^n + 6∑a[n]x^n

(1/x^2)∑ a[n+2]x^(n+2) = -(1/x)∑a[n+1]x^(n+1) + 6∑a[n]x^n

A(x) = ∑a[n]x^n

=> (1/x^2)(A(x) - a - a) = -(1/x)(A(x) - a) + 6A(x)

A(x) - a - a = -x(A(x) - a) + 6x^2A(x)

A(x)-2+1 = -xA(x) + 2x + 6x^2A(x)

A(x) +xA(x) - 6x^2A(x) = 2x + 1

A(x)(1+x-6x^2) = 2x+1

A(x) = (2x+1)/(1+x-6x^2)

and this functions series expansion does not have the right coefficients as mentioned before.. i really cant see where ive gone wrong..

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Stephen Tashi
A(x)(1+x-6x^2) = 2x+1
A(x) (1 + x - 6x^2) = 2x - 1

(1/x^2)∑ a[n+2]x^(n+2) = -(1/x)∑a[n+1]x^(n+1) + 6∑a[n]x^n

A(x) = ∑a[n]x^n

=> (1/x^2)(A(x) - a - a) = -(1/x)(A(x) - a) + 6A(x)
an x is missing from one of the terms above