Page 275 - Demo

269v 0,9; 0,81; 0,729; 0,6561; 0,59049vi 1,1; 1,21; 1,331; 1,4641; 1,61051 b i Kufizat afrohen te 1. ii Kufizat b%u00ebhen m%u00eb t%u00eb vogla.iii Kufizat b%u00ebhen m%u00eb t%u00eb vogla. iv Kufizat b%u00ebhen m%u00eb t%u00eb m%u00ebdha.v Kufizat b%u00ebhen m%u00eb t%u00eb vogla. vi Kufizat b%u00ebhen m%u00eb t%u00eb m%u00ebdha.9.3Z1 Ka shum%u00eb p%u00ebrgjigje t%u00eb mundshme duke p%u00ebrfshir%u00eb:a 4, 7, 10, 13, 16, 19 b 4, 8, 16, 32, 64, 128 c 4, 5, 9, 14, 23, 37 d 4, 6, 10, 16, 24, 342 a 32 b 7 c %u201319,5 d %u20131, 7e 768 f 1 g 9 3 h 163 a Po, kjo metod%u00eb krijon modelet e m%u00ebposhtme q%u00eb paraqesin numrat trek%u00ebndor%u00eb. x x x xx xx xxx b Numrat pes%u00ebk%u00ebndor%u00eb: 1, 5, 12, 22%u2026 Numrat gjasht%u00ebk%u00ebndor%u00eb: 1, 6, 15, 28%u20265 a Alternativ%u00ebn e tret%u00eb: N%u00eb fund t%u00eb shkurtit: 1 048 576 b Alternativ%u00ebn e par%u00eb: 5 000 lek%u00eb6 a Ky varg gjeometrik, ka term t%u00eb par%u00eb 12 dhe her%u00ebs 12b 12 , 34 , 78 , 1516, 3132c Katrori n%u00eb diagram ka syprin%u00eb 1. S(n)mbulon nj%u00eb pjes%u00eb t%u00eb katrorit dhe pjes%u00ebn tjet%u00ebr nuk e mbulon. S(n+1) do mbuloj%u00eb hap%u00ebsir%u00ebn e S(n) plus gjysm%u00ebn e hap%u00ebsir%u00ebs s%u00eb ngelur. Sa m%u00eb shum%u00eb rritet S(n), aq m%u00eb e vog%u00ebl b%u00ebhet pjesa q%u00eb nuk mbulohet, por nuk zhduket asnj%u00ebher%u00eb, pra termat e S(n) i afrohen pafund%u00ebsisht 1, por nuk e arrijn%u00eb kurr%u00eb at%u00eb.7 a N%u00eb hapin e dyt%u00eb, numrin e shum%u00ebzojm%u00eb me 7.b N%u00eb hapin e dyt%u00eb, numrin e shum%u00ebzojm%u00eb me 5.c P%u00ebrgjigjet e nx%u00ebn%u00ebsve.8 a ab, ab3, ab5, ab7, ab9b c2d7, c4d5, c6d3, c8d, c10d%u22121c 3x %u2212 6, 6x2 %u2212 12x, 12x3 %u2212 24x2, 24x4 %u2212 48x3, 48x5 %u2212 96x49 N%u00eb t%u00eb gjitha rastet, supozojm%u00eb az0. N%u00ebse r > 1 ose r < %u22121, kufizat rriten pambarimisht. N%u00ebse r = 1, kufizat nuk ndryshojn%u00eb. N%u00ebse r = %u22121, kufizat do t%u00eb ndryshojn%u00eb shenj%u00eb a, -a, a, -a...... N%u00ebse %u22121 < r < 1, kufizat do t%u00eb konvergjojn%u00eb.P%u00ebrs%u00ebritje 93 a 6n %u2212 5 b 7n + 8 c 63 %u2212 12n d %u22121,5n %u2013 54 a 175 b 3405 a n2 + 3 b 3n2 %u2212 n c 2n2 + 3n %u2212 16 a progresion gjeometrik b varg kuadratik c varg Fibona%u00e7id progresion aritmetik e varg kuadratik f progresion gjeometrik7 a i 15, 21 ii 125, 136 b i 12 n(n + 1) ii 21n8 a 32, 64, 128 b 2n9 a 2 2n %u2212 2; 19 2 b n2n + 1; 1021Vler%u00ebsim 92 b 10n %u2013 93 a i 2n + 8 ii 2n + 2 b 54 c 424 a %u00c7do kufiz%u00eb %u00ebsht%u00eb shum%u00eb e dy termave paraardh%u00ebs. b 55, 89 c 144 = 122 d Fibona%u00e7i5 a E sakt%u00eb. 2 %u00d7 10 + 7 = 27 b E gabuar. 6 %u00d7 1 %u2212 5 = 1, 6 %u00d7 2 %u2212 5 = 7, 6 %u00d7 3 %u2212 5 = 13 c E gabuar. 13 %u2212 3 %u00d7 100 = %u2212287 d E gabuar. 102 %u2212 10 = 100 %u2212 10 = 90 e E sakt%u00eb. 15 %u2212 3 %u00d7 1002 = 15 %u2013 30 000 = %u221229 9856 a Jo. Raporti zvog%u00eblohet: 1 : 2; 1 : 4; 1 : 6.7 a Deni b D = t (t + 1)/ 2 c i 55 pika ii 1 275 pika iii 5 050 pika