wiad ciągi: Wyznacz wzór ogólny ciągu danego rekurencyjnie a₀=a₁=a₂=a₃=0 i a_n+4+a_n+3+2a_n+2+a_n+1+a_n=12n

Mariusz: A(x)=∑_n=0^∞a_nxⁿ ∑_n=0^∞(a_n+4+a_n+3+2a_n+2+a_n+1+a₁)xⁿ=∑_n=0^∞12nxⁿ ∑_n=0^∞a_n+4xⁿ+∑_n=0^∞a_n+3xⁿ+2(∑_n=0^∞a_n+2xⁿ) +∑_n=0^∞a_n+1xⁿ+∑_n=0^∞a_nxⁿ=12x(∑_n=0^∞nxⁿ⁻¹)\\

1		1
	(∑n=0∞an+4xn+4)+		(∑n=0∞an+3xn+3)+
x4		x3

2		1
	(∑n=0∞an+2xn+2)+		(∑n=0∞an+1xn+1)+∑n=0∞anxn
x2		x

	d
=12x		(∑n=0∞xn)
	dx

∑_n=0^∞a_n+4xⁿ⁺⁴+x(∑_n=0^∞a_n+3xⁿ⁺³)+2x²(∑_n=0^∞a_n+2xⁿ⁺²)

	d		1
+x3(∑n=0∞an+1xn+1)+x4(∑n=0∞anxn)=12x5		(		)
	dx		(1−x)

(∑_n=0^∞a_nxⁿ−a₀−a₁x−a₂x²−a₃x³) +x(∑_n=0^∞a_nxⁿ−a₀−a₁x−a₂x²)+2x²(∑_n=0^∞a_nxⁿ}−a₀−a₁x)

	(−1)
+x3(∑n=0∞an+1xn+1−a0)+x4(∑n=0∞anxn) = 12x5		(−1)
	(1−x)2

	12x5
A(x)(1+x+2x2+x3+x4)=
	(1−x)2

	12x5
A(x)=
	(1−x)2(1+x+2x2+x3+x4)

	12x5
A(x)=
	(1−x)2((1+x+x2 + x2 + x3+x4))

	12x5
A(x)=
	(1−x)2((1+x+x2)+x2(1+x+x2))

	12x5
A(x)=
	(1−x)2(1+x2)(1+x+x2)

	π		8√3		2π		π
an=−6+2(n+1)+6sin(		n)+		cos(		n+		)
	2		3		3		6

	π		8√3		π
an=2n−4+6sin(		n)+		cos(		(4n+1))
	2		3		6