suma szeregu Mateusz: Oblicz sumę:

	(2n+3n)
∑(n=0 do ∞)
	(n+2)*4n

jc: ln(1−x) = − [x + x²/2 + x³/3 + x⁴/4 + ...] Rozdzielasz na dwa szeregi. 1/2 + (1/2)² /3 + (1/2)³/4 + ... = 2 [(1/2) + (1/2)²/2 + (1/2)³/3 + ...] − 1 = − 2 ln(1−1/2) − 1 Drugi szereg podobnie. W sumie − 2 ln(1−1/2) − (4/3) ln(1−3/4) − 2 = 2 ln 2 + 4/3 ln 4 − 2 = (14/3) ln 2 − 2, sprawdź!

piotr:

1		xn+2		1		1
	∑n=0∞		=		∫∑n=0∞ xn+1dx =		∫x∑n=0∞ xndx
x2		n+2		x2		x2

Adamm:

	1
dla x≠1, ∑n=0∞xn =
	1−x

	1
∫0x∑n=0∞xndx = ∫0x		dx
	1−x

	1
∑n=0∞		xn+1=−ln|1−x|
	n+1

	1
∑n=0∞		xn+1=−ln|1−x|
	n+1

	1
x+∑n=0∞		xn+2=−ln|1−x|
	n+2

	1		−ln|1−x|−x
∑n=0∞		xn=
	n+2		x2

dla x=1/2 mamy 4ln2−2 dla x=3/4 mamy

32		4
	ln2−
9		3

w sumie

68		10
	ln2−
9		3

Mila:

	2n		3n
=∑(n=0 do ∞)		+∑(n=0 do ∞)		=
	(n+2)*4n		(n+2)*4n

	1   ( )n   2		3   ( )n   4
=∑(n=0 do∞)		+∑(n=0 do ∞)		=
	n+2		n+2

	1   ( )n+2   2		16		3   ( )n+2   4
=4*∑(n=0 do∞)		+		*∑(n=0 do∞)		=
	n+2		9		n+2

	1   ( )n   2		1		16		3   ( )n   4		3
4*[(∑(n=1do∞)		)−		]+		*[(∑(n=1do∞)		)−		]=
	n		2		9		n		4

	1		16		3		16		3
=4ln(1−		)−1−2+		ln(1−		)−1−		*		=
	2		9		4		9		4

	68		10
=		ln(2)−
	9		3

piotr:

1		1		x dx		−x−ln(1−x)
	∫x∑n=0∞ xn dx =		∫		=
x2		x2		1−x		x2

dla x=1/2 −2+ln(16) x=3/4

16		4
	(−3/4+ln(4)) =		(ln(256)−3)
9		9

i razem mamy:

	4
−2+ln(16) +		(ln(256)−3)
	9