SS baa: Oblicz 1*3+3*5+5*7+...+999*1001

Blee: Masz do obliczenia sume ciagu o wyrazie ogolnym: a_n = (2n−1)(2n+1)

Mariusz: S₀=0 S_n=S_n−1+(2n−1)(2n+1) (4n²−1) G(x)=∑_n=0^∞S_nxⁿ ∑_n=1^∞S_nxⁿ=∑_n=1^∞S_n−1xⁿ+∑_n=1^∞(4n²−1)xⁿ

	1
∑n=0∞xn=
	1−x

d		d	1
	(∑n=0∞xn)=
dx		dx	1−x

	1
∑n=0∞nxn−1=−		(−1)
	(1−x)2

	1
∑n=1∞nxn−1=
	(1−x)2

	1
∑n=0∞(n+1)xn=
	(1−x)2

d		d		1
	(∑n=0∞(n+1)xn)=		(		)
dx		dx		(1−x)2

	2
∑n=0∞(n+1)nxn−1=−		(−1)
	(1−x)3

	2
∑n=1∞(n+1)nxn−1=
	(1−x)3

	2
∑n=0∞(n+2)(n+1)xn=
	(1−x)3

d		d		2
	(∑n=0∞(n+2)(n+1)xn)=		(		)
dx		dx		(1−x)3

	6
∑n=0∞(n+2)(n+1)nxn−1=−		(−1)
	(1−x)4

	6
∑n=1∞(n+2)(n+1)nxn−1=−
	(1−x)4

	6
∑n=0∞(n+3)(n+2)(n+1)xn=−
	(1−x)4

4n²−1=4(n²+3n+2)−12(n+1)+3 ∑_n=0^∞S_nxⁿ=x(∑_n=1^∞S_n−1xⁿ⁻¹)+∑_n=0^∞(4n²−1)xⁿ−(−1)

	8		12		1
∑n=0∞Snxn=x(∑n=0∞Snxn)+		−		−		+1
	(1−x)3		(1−x)2		1−x

	8		4		3
G(x)=xG(x)+		−		+		+1
	(1−x)3		(1−x)2		1−x

	8		12		3
G(x)(1−x)=		−		+		+1
	(1−x)3		(1−x)2		1−x

	8		12		3		1
G(x)=		−		+		+
	(1−x)4		(1−x)3		(1−x)2		1−x

	4
Sn=		(n+3)(n+2)(n+1)−6(n+2)(n+1)+3(n+1)+1
	3

Mila:

	n*(n+1)*(2n+1)
1) Sn=∑(k=1 do n) k2=
	6

2) S=1*3+3*5+5*7+...+999*1001 a_n=(2n−1)*(2n+1) a_n=4n²−1 a_n=999*1001⇔ 2n−1=999 i 2n+1=1001⇔n=500 S₅₀₀=∑(n=1 do 500) (4n²−1)=4*∑(n=1 do 500) n²−500=

	500*501*1001
=4*		−500
	6

oblicz

baa: Dzięki, zrobiłem błąd obliczeniowy po drodze i mi nie wychodziło