Rozwiąż równanie Mirek: Rozwiąż równanie (prosiłbym wraz z obliczeniami)

	1
fn+1 = 4fn +
	2n

Adamm:

	1
g(x) = ∑0 fn xn = f0 + x ∑0 (4fn+1/2n) xn = f0 + x (4g(x) +		)
	1−x/2

	x
g(x) = f0 + 4xg(x) +
	1−x/2

	x
(1−4x)g(x) = f0 +
	1−x/2

	f0		x
g(x) =		+
	1−4x		(1−x/2)(1−4x)

x		A		B
	=		+
(1−x/2)(1−4x)		1−x/2		1−4x

x = A(1−4x)+B(1−x/2) 2 = −7A, x = 2 A = −2/7 B = −A = 2/7

	f0		−2/7		2/7
g(x) =		+		+		= ∑0 (f0*4n+(−2/7)(1/2)n+(2/7)4n)xn
	1−4x		1−x/2		1−4x

f_n = (f₀+2/7)*4ⁿ + (−2/7)*(1/2)ⁿ sprawdźmy (f₀+2/7)*4⁰ + (−2/7)*(1/2)⁰ = f₀ 4*[(f₀+2/7)*4ⁿ + (−2/7)*(1/2)ⁿ]+1/2ⁿ = (f₀+2/7)*4ⁿ⁺¹ + (−8/7)*(1/2)ⁿ + 1/2ⁿ = (f₀+2/7)*4ⁿ⁺¹ + (−1/7)*(1/2)ⁿ = (f₀+2/7)*4ⁿ⁺¹ + (−2/7)*(1/2)ⁿ⁺¹

kerajs: Inaczej

	1
fn=4nf0+∑i=1 n 4n−i
	2i

f_n=4ⁿf₀+∑_i=1 ⁿ 2²ⁿ⁻³ⁱ

	1   22n−3(1−( )n)   23
fn=4nf0+
	1   1−   23

	1   22n(1− )   23n
fn=4nf0+
	7