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2010-01-29T20:35:05+01:00
1 +2 +3 + ...+ n = [n*(n+1)]/2
zatem
an = [n²+3n -2]/(1 +2 + ...+n) = 2*[n² + 3n -2]/[n*(n+1)] =
=[2n² + 6n - 4]/[n² + n] = [2 + 6/n -4/n²]/[1 + 1/n]
lim { [2+6/n - 4/n²]/[1 +1/n]} =
n--> ∞
=[ lim (2 +6/n -4/n²)] :
n --> ∞
: [ lim (1 +1/n)] = 2 : 1 = 2,
n--> ∞
lim (6/n) = 0
n--> ∞
lim ( 4/n²) = 0
n --> ∞
lim (1/n) = 0
n --> ∞

a1 = [2*1² +6*1 - 4]/[1² +1] = 4/2 = 2
a2 = [2*2² +6*2 - 4]/[2² + 2]= 16/6 = 8/3
a3 = [2*3² +6*3 -4]/[3²+3] = 32/12= 8/3
S3 = a1 +a2 +a3 = 2 + 8/3 + 8/3 = 6/3 + 16/3 = 22/3

a6 = [2*6² +6*6 - 4]/[6² + 6] = [72 +36 - 4]/42 = 104/42 = 52/21
Ciąg nie jest rosnący ani malejący.