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n²+n-an=0
Δ=1+4an
n=(-1+√(1+4an))/2 bo n>0
an+1 = (n+1)(n+2) = n²+3n+2 = [(-1+√(1+4an))/2]² +3∙(-1+√(1+4an))/2 +2 =
1/4 -1/2√(1+4an) +1/4 ∙(1+4an) -3/2 +3/2√(1+4an) +2 =
an + √(1+4an) +1
Wzór rekurencyjny:
an+1=(n+1)((n+1)+1)
an+1=(n+1)(n+2)
an+1=+2n+n+2
an+1=+3n+2