zad MD 2015 I 7

(Fn)
Fn = Fn-1 + Fn-2, F1 = F2 = 1.
1 1 1
1 2 3 = (3 - 2)(3 - 1)(2 - 1).
2 2 2
1 2 3
an = 2an-1 - n a0 = 3
an = an-1 + 2an-2 + 3n a0 = 0 a1 = 1
an+1 = 3an + 2n + 1 a0 = 0
an = 2an-1 + 2n + 1
bn = 3bn-1 + 5n2
cn = 3cn-1 - 2cn-2 + 2n + n + 1
sn = 5sn-1 - 6sn-2 + 2nn.
2n(An2 + Bn + C)

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